2018-G12-Math-E

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab2.2 FINDING f’(x) FROM DEFINITION OF DERIVATIVEGiven a function f , f ' ( x) if it exists, can be found by the following four steps Find f ( x + d x)Step I Simplify f ( x + d x) - f ( x)Step IIStep III Divide f ( x + d x) - f ( x) by d x to get f (x +dx)- f (x) and simplify itStep IV (x +d x)- (x) dx Find lim f dx f d x→0 The method of inding derivatives by this process is called diferentiation by deinitionor by ab-initio or from irst principle.Example 1: Find the derivative of the following functions by deinition==(a) f ( x) c (b) f ( x) x2Solution: (a) For f ( x) = c (i) f ( x + d x) =c (ii) f ( x + d x) - f ( x) = c - c = 0(iii) f (x +d x)- f ( x=) d=0x 0 dx (x + d x) -=f ( x)(iv) lim f dl=xi→m0 ( 0 ) 0 dx d x→0 Thus f '(x) = 0 , that is, d (c) =0(b) For f(x) = x2 dx(i) f ( x + d x) =( x + d x)2(ii) f ( x + d x) - f ( x) = ( x + d x)2 - x2 = x2 + 2xd x + (d x)2 - x2 = 2xd x + (d x)2 =(2x + d x)d x version: 1.1 7

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab (iii) f (x +d x) - f (x) =(2+x +ddxx)d x =≠2x d x , (d x 0) eLearn.Punjab version: 1.1 dx (x +d x) - ( x=) (iv) lim f f lim ( 2 x + d x=) 2x dx d x→0 d x→0 i.e., f ' ( x) = 2xExample 2: Find the derivative of x at x = a from irst principle.Solution: If f ( x) = x , then(i) f ( x + d x) = x + d x and(ii) f ( x + d x) - f ( x) = x + d x - x ( )(= )x x+dx - x x+dx +  rationalizing the  x+dx + x numerator = (x +d x)- x x+dx + x dxi.e., f (x +d x) - f (x) =x + dx + x (I)(iii) Dividing both sides of(1)by d x , we have =f=( x + ddxx) - f ( x) d ≠ dx x) 1 (d x 0) x( x+dx + x+dx + x(iv) Taking limit of both the sides as d x → 0, we have lim f (x +d x)- f (x) = lim  1 x  x+dx + d x→0 dx d x→0i.e., =f ' ( x) =x +1 x 1 (x > 0) 2xand f '(a) = 1 2a 8

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjabor eLearn.Punjab version: 1.1Puttin=g x a==in f ( x) x , gives f (a) aSo f ( x) - f (a) = x - aUsing alternative form for the deinition of a derivative, we have f ( x) - f (a ) = x- a - a x-a x ( ()( ) )= x- a x+ a (rationalizing the numerator) (x - a) x+ a( )==( x -≠a x - a 1 (x a) (II) x x+ a ) + aTaking limit of both the sides of (II)as x → a, gives lim f=(=xx) -- af (a) lim 1 a 1 x+ a+ a x→a x→ai.e., f '(a) = 1 2aExample 3: If y= 1 , then find dy at x= - 1 by ab-initio method. x2 dxSolution: Here y= 1 , so (i) x2 (ii) y + d y = ( x 1 x)2 +dSubtracting (i) from (ii), we get=d y ( x 1 x )2 =- x12 x2 - ( x + d x)2 +d x2 ( x + d x)2 ( )( )) = x + ( x +d x) x -(x + d x x2 (x + d x)2 9

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab== ( 2x + d x)( -d x ) -d x(2x + d x) (iii) x x2 ( x + d x)2 x2 ( +d x)2 Dividing both sides of (iii) by d x,, we have ==dd xy -d x(2x x+≠)d2 x ) -(2x +d x) (d x 0) x2 (x +d d x2 ( x + d x)2 x Taking limit as d x → 0,, gives lim d y = lim -(2x +d x) d x x2 ( x + d x)2 d x→0 d x→0 = -(2x) (Using quotient theorem of limits) ( )x2 x2 =( --12)3= -2 -2 i.e., d=y x3 and d=y |x= -1 -1 2 dx dxNote: The value of dy at x = -1 is written as dy | . dx dx x=-1Example 4: 2 Find the derivative of x3 and also calculate the value of derivative at x = 8.Solution: Let f (x) = 2 .Then x3 f ( x + d x) =( x + d )2 x3 and  ( x + d x )2 2  ( x+ d )4 + ( x+ d x ) 2 .x 2 + x 4  3 x )4 x 3 3 3 ( - x3 +( x3 x) 2 2f (x +d x)- f ( x) = ( x + d x ) 2 - 2 = 3 +d 3 4 3 x+d .x 3 x3 + x3 version: 1.1 10

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab ( x + d x ) 2 3 -  x 2 3 ( x + d x)2 - x2 eLearn.Punjab 3  3 version: 1.1 ( x + d )4 + ( x + d )2 2 + 4 (x + d )4 + (x +d )2 2 + 4 x3 x3 .x 3 x3 x3 x3 .x 3 x3i.e., f (x +d x) - f (x) = d x(2x +d x) 2 4 (i) 3 (x+ d x )4 + ( x + d x ) 2 .x + x3 3 3 Dividing both the sides of (i) by d x , we get f (x +d x)- f (x) = 2x +d x (ii) dx (x + d x ) 4 + ( x + d x ) 2 .x 2 + 4 3 3 3 x3 Taking limit of both the sides as d x → 0, we have f '(=x) 4 22=x 2 =4 2x 2 4 1 x3 + x3.x3 + x3 3x3 3x3and =f '(8) =2 1 1 3.(8)3 3Example 5: Find the derivative of x3 + 2x + 3 .Solution: Let y = x3 + 2x + 3. Then (i) y + d y =( x + d x)3 + 2( x + d x) + 3 (ii) d y = ( x + d x)3 + 2( x + d x) + 3 - x3 + 2x + 3 = ( x + d x)3 - x3  + 2 ( x + d x) - x + (3 - 3) = ( x + d x) - x ( x + d x)2 + ( x + d x) x + x2  + 2d x (iii) d y = d x ( x + d x)2 + ( x+d x) x + x2  + 2d x d x dx 11

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab = ( x + d x)2 + ( x + d x) x + x2 + 2(iv) lim =dd xy lim ( x + d x)2 + (x + d x)x + x2 + 2 d x→0 d x→0 dy = (x)2 + ( x) x + x2 + 2 dx( )i.e., d x3 + 2x + 3 = 3x2 + 2 dx2.2.1 Derivation of xn where ndZ.(a) We ind the derivative of xn when n is positive integer. (a) Let y = xn . Then y + d y =( x + d x)nand d y =( x + d x)n - xnUsing the binomial theorem, we have  x n n(n -1) (d x)n - 2 ( )d y= + nxn-1.d x + xn-2 (d x2+) .+.. xni.e.,=d y d x nxn-1 + n(n -1) x+n-2.+d x ... (d x)n-1  (i) 2Dividing both sides of (i) by d x , gives d y= nxn-1 + n( n -1) xn-2 . d+x +... ( d x )n-1 (ii) d x 2Note that each term on the right hand side of (ii) involves d x except the irst term, so taking the limit as d x → 0 , we get dy = nxn-1 dx( )==As y xn , so d xn n.xn-1 dx version: 1.1 12

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.PunjabNote: ( ) ( )If n = 0, then the formula d xn = nxn-1 reduces to d =x0 0=x0-1 0 i.e., dx dxd (1) = 0 which is correct by example 1 part (a).dx(b) Let y = xn where n is a negative integer.Let n = -m (m is a positive integer). Then =y x=-m 1 (i) xm (ii)and y + d y =( x +1d x)mSubtracting (i) from (ii). gives =d y ( x 1 x)m =- x1m xm - ( x + d x)m +d xm ( x + d x)m xm -  xm + mxm-1d x + m(m -1) xm-2 (d x)2 + ... + (d x)m  = 2 xm ( x + d x)m (expanding ( x + d x)m by binomial theorem) -d x  mxm-1 + m(m -1) x dm-2 x + ... + (d )x m-1  = 2 xm .( x + d x)m=and dd xy x m+( -1 x)m . mxm-1 + m(+m -1) xm-2 .d x ... (d x)m-1  x+d 2 Taking limit when d x → 0 , we get ( )dy -1 (all terms containing d x ,vanish) xm.xm dx = mxm-1 version: 1.1 13

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab ( ) ( )= -m xm-1.x-2m = -m x(-m)-1 =nxn-1 [ m- n=] or d ( x)n = nxn-1 dxSo far we have proved that d [ x]n = nxn-1, if n ∈ Z dx The above rule holds if n ∈Q - ZFor example d =x 23  2=x 23-1 2 dx 3 1 3x3The proof of d  x n  = nxn-1 when n∈Q - Z is left as an exercise. dxNote that d  xn  = nxn-1 is called power rule. dx Exercise 2.11. Find by deinition, the derivatives w.r.t ‘x’ of the following functions deined as:(i) 2x2 + 1 (ii) 2 - x (iii) 1 (iv) 1 (v) 1 x x3 x-a(vi) x( x - 3) (vii) 2 (viii) (x + )1 3 5 x4 43 (ix) x2 (x) x 2(xi) xm,m∈ N (xii) 1 (xiii) x40 (xiv) x-100 xm,m∈ N2. Find dy from irst principle if dx(i) x + 2 (ii) 1 x+a version: 1.1 14

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab2.2.2 DIFFERENTIATION OF EXPRESSIONS OF THE TYPES: (ax + b)n and 1 , n =1,2,3... (ax + b)n We ind the derivatives of (ax + b)n and 1 from the irst principle when n ∈ N (ax + b)nExample 1: Find from deinition the diferential coeicient of (ax + b)n w.r.t. ‘ x ‘ when nis a positive integer.Solution: Let y = (ax + b)n, (n is a positive integer) Then y + d y= a( x + d x) + bn= (ax + b) + ad xn Using the binomial theorem we havey + d y =+(ax +b)n 1n +(ax b)n-1 (ad+x)  n  +( ax )b n-2 (ad+x)2 + ... (ad x)n 2dy= (y +d y)- y = 1n ( ax + )b n-1 ( ad x ) +  n  ( ax + b )n-2 .a2 (d x )2 + ... + an (d x )n 2= d x 1n  ( ax + b )n-1 .a +  n  ( ax + b )n-2 .a 2d x + ... + a n (d x )n-1  2So =dd xy 1n  ( ax + b )n-1 a +  n  ( ax + b )n-2 .a 2d x + ... + a n (d x )n-1 2Taking limit when d x → 0 , we have =dlixm→0 dd xy lim 1n  ( ax + b )n-1 .a +  n  ( ax + b )n-2 .a2d x + ... + an (d x )n-1  2 d x→0O=r dy 1n (ax + b)n-1.a [All other terms tends to zero when d x → 0 ] dx Thus d (ax + b)n = n(ax + b)n-1.a dx version: 1.1 15

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.PunjabExample 2: Find from irst principle, the derivative of 1 w.r.t. ‘ x ’, (ax + b)nSolution: Let y= 1 (when n is a positive integer). Then (ax + b)n y + d y =a ( + 1 ) +  n and dx x b d y = y + d y - y = ( ax + 1 ad xn - ( ax 1 b)n + b) +or d y = (ax + b)n - (ax + b + ad x)n (ax + b) + ad xn (ax + b)nor dy ( + + b+) + -1 -n ( + b)n x ( ax b) ad xn ( ax b )n ] (I) ax ad x ax + +Using the binomial theorem, we simplify the expression (ax + b) + ad xn - (ax + b)n ,That is, (ax + b) + ad xn - (ax + b)n= [(ax + b)n + 1n (ax + )b n-1 (ad x) +  n  ( ax + )b n-2 .a2 (d x)2 + ... + ( ad x)n ] 2= 1n  ( ax + )b n-1 .ad x +  n  ( ax + b )n-2 .a2 (d x )2 + ... + an (d x )n 2= d x 1n  ( ax + b )n-1 .a +  n  ( ax + b )n-2 a 2d x + ... + a n (d x )n-1  2Now (I) becomesd y =(ax + b) + addxxn (ax + b)n [1n (a-x )b n-1.a version: 1.1 16

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab +  n  ( ax + b )n-2 .a 2d x + ... + a n (d x )n-1 ] 2and d y =(ax + b) + ad1xn (ax + b)n [1n (a-x )b n-1 .a + d x +  n  ( ax + b )n-2 .a 2d x + ... + a n (d x )n-1 ] 2Using the product and sum rules of limits when d x → 0 , we have dy =(ax + b)n1(ax + b)n .1n (a-x )b n-1 .a alldlixom→th0 dderxy+te=rmddyxs and  dx containing d x vanish or d = ( ax 1 b)n  =- (ax-+n+ba)n+1 =n ( ax )b -(n+1) .a dx + Exercise 2.21. Find from irst principles, the derivatives of the following expressions w.r.t. their respective independent variables: (i) (ax + b)3 (ii) (2x + 3)5 (iii) (3t + )2 -2 (iv) 1 1 (ax + b)5 (v) (az - b)7 version: 1.1 17

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab2.3 THEOREMS ON DIFFERENTIATION We have, so far proved the following two formulas:1. dy (c) = 0 i.e.. the derivative of a constant function is zero. dx( )2. d xn = nxn-1 power formula (or rule) when n is any rational dx number. Now we will prove other important formulas (or rules) which are used to determinederivatives of diferent functions eiciently. Henceforth, in all subsequent discussion, f, g, hetc. all denote functions diferentiable at x, unless stated otherwise.3. Derivative of y = cf ( x)Proof: Let y = cf ( x) . Then(i) y + d y= cf ( x + d x) and(ii) y + d y - =y cf ( x + d x) - cf ( x)or d y= c | f ( x + d x) - f ( x) | (factoring out c)(iii) d y = c  f (x +d x)- f ( x)  d x dx Taking limit when d x → 0(iv) lim d y lim c. f (x +d x)- f ( x)  c. lim f (x +d x)- f (x) d x d x→0 d x→0 d x→0 dx dxA constant factor can be taken out from a limit sign.Thus dy =c f '( x) ,that is, c f ( x)' = cf ' ( x) dxor dy = cf '( x) = c f ( x)' = cf ' ( x) dx version: 1.1 18

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.PunjabExample 1: Calculate d  3x 4  dx 3Solution: d  3x 4  = 3 d  x 4  (Using Formula 3) dx 3 dx 3 = 3=x 4 x 43-1 1 (Using power rule) 3 4x34. Derivative of a sum or a Diference of Functions: If f and g are diferentiable at x , then f + g, f - g are also diferentiable at xand  f ( x) + g ( x)' =f '( x) + g '( x) , that is, d  f ( x) + g ( x=) d  f ( x ) + d  g ( x ) Also dx dx dx f ( x) - g ( x)' =f '(x) - g '(x). that is, d  f ( x) - g ( x=) d  f ( x ) - d  g ( x ) dx dx dxProof: Let f=( x) f ( x) + g ( x) . Then(i) f ( x + d x) = f ( x + d x) + g ( x + d x) and(ii) f ( x + d x) - f ( x) = f ( x + d x) + g ( x + d x) -  f ( x) + g ( x) =  f ( x + d x) - f ( x) + g ( x + d x) - g ( x) (rearranging the terms)(iii) f=( x + ddxx) - f ( x) +f ( x + d x) - f (x) g(x +d x)- g(x) dx dx Taking the limit when d x → 0(iv) =dlixm→0 f ( x + ddxx) - f ( x) +dlixm→0  f ( x + d x) - f (x) g ( x + d x) - g ( x )  dx dx= +dlixm→0 f (x +d x)- f (x) lim g ( x + d x) - g ( x) dx d x→0 dx (The limit of a sum is the sum of the limits) =f ' x f '( x) + g '( x) , that is  f ( x) + g ( x)' = f '( x) + g '( x)or d  f ( x ) + g ( x=) d  f ( x) + d  g ( x) dx dx dxThe proof for the second part is similar. version: 1.1 19

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.PunjabNote: Sum or diference formula can be extended to ind derivative of more than twofunctions.Example 1: Find the derivative of y = 3 x4 + 2 x3 + 1 x2 + 2x + 5 w.r.t. x . 432Solution: y = 3 x4 + 2 x3 + 1 x2 + 2x + 5 432 Diferentiating with respect to x, we havedy  3 x4 + 2 x3 + 1 x2 + 2x=+ 5 d  3 x4  + d  2 x3  + d  1 x2  + d (2x) + d (5)dx 4 3 2 dx 4 dx 3 dx 2 dx dx (Using formula 4)= 3 d ( x4 ) + 2 d ( x3 ) + 1 d ( x2 ) + 2 d ( x ) + 0 (Using formula 3 and 1) 4 dx 3 dx 2 dx dx (By power formula)( ) ( ) ( ) ( )= 3 4x4-1 + 2 3x3-1 + 1 2x2-1 + 2 1.x1-1 4 32= 3x3 + 2x2 + x + 2Example 2: ( )( )Find the derivative of y =x2 + 5 x3 + 7 with respect to x.Solution: y =( x2 + 5)( x3 + 7) =x5 + 5x3 + 7x2 + 35 Diferentiating with respect to x, we get dy= d  x5 + 5x3 + 7x2 + 35 dx dx ( ) ( )=d dx  x5  + 5 d x3 +7 d x2 + d [35] (Using formulas 3 and 4) dx dx dx = 5x 5-1 + 5 x 3x 3-1 + 7 x 2x 2-1 + 0 = 5x4 + 15x2 + 14x version: 1.1 20

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.PunjabExample 3: ( )( )Find the derivative of y = 2 x + 2 x - x with respect to x.( )( )Solution: y = 2 x + 2 x - x = 2( )x +1 . x ( )x=-1 2 x ( x +1)( x -1) =2 x ( x +1=) 2 3 - 1  x2 x2Diferentiating with respect to x , we have=dy d 2 3 - 1  dx dx x2 x2 = 2  d  3  - d  1  = 2  3 3 - 1 - 1 1 - 1 dx dx 2 2  x2 x2 x2 x2 1 -1 x- 1 = 3x -1 = 3x2 - x 2 = 3 xx5. Derivative of a product. (The product Rule)If f and g are diferentiable at x, then fg is also diferentiable at x and f (=x) g ( x)' f '( x) g ( x) + f ( x) g '( x) , that is,=ddx  f ( x) g ( x) + d  f ( x) g ( x) f ( x )  d  g ( x ) dx dxProof: Let f ( x) = f ( x) g ( x). Then(i) f ( x + d x) = f ( x + d x) g ( x + d x)(ii) f ( x + d x) - f ( x) = f ( x + d x) g ( x + d x) - f ( x) g ( x)Subtracting and adding f ( x) g ( x + d x) in step (ii), givesf(x +d x)-f(x) = f (x +d x)g(x +d x)- f (x)g(x +d x)+ f (x)g(x +d x)- f (x)g(x) =  f ( x + d x) - f ( x) g ( x + d x) + f ( x) g ( x + d x) - g ( x) version: 1.1 21

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab(iii) =f ( x + ddxx) - f ( x)  f (+x + ddx+x) - f ( x)  g ( x d x) f ( x )  g ( x + d x) - g ( x)  eLearn.Punjab version: 1.1Taking limit when d x → 0 dx(iv) f (x +d x) -f (x) lim dx d x→0 = lim  f (x +d x) - f ( x).g ( x +d x)+ f (x ). g(x +d x)- g d( xx))- = x ) . lim g (x +d x)+ + d x→0 dx f( lim f dx dx d x→0 f (x +d x)- d x→0 ( x ). lim g ( x g ( x ) lim d x→0 dx d x→0 (Using limit theorems)Th=us f '( x) f '( x) g ( x) + f ( x) g '( x)  lim g ( x + d x ) =g ( x) d x→0or dd=x  f ( x).g ( x) +d  f ( x ) .g ( x ) f ( x)  d g ( x ) dx dx( )( )Example: Find derivative of y = 2 x + 2 x - x with respect to x( )( )Solution: y = 2 x + 2 x - x = 2( x +1)( x - x ) Diferentiating with respect to x, we get ( )( )=dy d dx dx 2  x +1 x - x  ( ) ( ) ( ) ( )= x  2 d x +1  x - x + x +1 d x - dx dx ( ) ( )=2  1 x 1 - 1 + 0  x- x+ x +1 × 1 - 1 x 1 - 1  2 2 2 2 22

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab ( ) ( )=2  2 1 x- x + x +1 x 1 - 1  x 2x ( )=2  x- x + x +1  2 x -1  2 x x 2 = 1  x - x + 2x - x + 2 x -1 x = 3x -1 x6. Derivative of a Quotient (The Quotient Rule) If f and g are diferentiable at x and g( x) ≠ 0 , for any x ∈ D( g ) then f is diferentiable gat x and  f ( x) ' = f '(x)g(x) - f (x)g '(x) g ( x) g ( x)2 d  f ( x)   d  f ( x) g ( x) - f ( x)  d g ( x) dx g ( x) dx dxthat is, = g ( x)2Proof: Letf ( x) = f (x) Then g(x) =gf ((xx x)(i) f ( x + d x) + d x) + d(ii) f ( x + d x) -=f ( x) f (x +d x) - g=f ((xx)) f (x +d x)g(x)- f (x)g(x +d x) g (x +d x) g(x)g(x +dx)Subtracting and adding f ( x) g ( x) in the numerator of step (ii), givesf(x +d x) -f ( x) =f ( x + d x) g (x) - f (x)g(x)- f (x)g(x +d x) + f (x)g(x) g(x)g(x +dx) = 1 +d x) ( f (x +d x) - f (x))g(x) - f (x)(g(x +d x) - g ( x)) g(x)g(x version: 1.1 23

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab(iii) f(x +dx)-f(x) - 1 x)  f (x +d x)- f ( x).g (x) f ( x ). g ( x + d x) - g ( x)  g dx ( x) g (x +d dx dx Taking limit when d x → 0(iv) lim f ( x + d x) - f ( x ) d x→0 dxlim  1 + d x)  f (x +d x) - f ( x).g ( x) - f (x). g(x +d x) - g(x) x→0 g ( x ) g (x dx dxUsing limit theorems, we have( )f '(x) g ( x-)1.g ( x)  f '( x) g (=x) f ( x) g+'( x)  lim g ( x d x) g(x) d x→0Thus  f ( x) ' f '(x)g(x) - f (x)g '(x) d  f ( x)  d  f ( x) g ( x) - f ( x)  d  g ( x)  g ( x) g ( x)2 dx g ( x)  dx  dx  or g ( x)2First Alternative Proof: f(x) = f (x) can be written as f (x) =f(x)g(x) g(x) Using the procedure used to prove product rule, quotient rule can be proved.Second Alternative Proof: We irst prove the reciprocal rule and then use product rule toprove the quotient rule.The reciprocal rule. If g is diferentiable at x and g ( x) ≠ 0 , then 1 is diferentiable at x and gd  1  = - d  g ( x )dx dx (x) g ( x)2 (Proof of reciprocal rule is left as an exercise) g version: 1.1 24

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab Using the product rule to f ( x ). g 1 ) , we have eLearn.Punjab version: 1.1 (x =ddx  f ( x). g (1x)  + d  f ( x ) . g 1 f ( x). d  g 1  dx dx (x) (x) = d  f (x ) + f ( x) - d g ( x) dx g x) ( x)2 ( dx  g d  f (x)   d  f ( x )  g (x)- f ( x)  d  g ( x )  dx g (x) dx g ( x)2 dxi.e., = ( )Find dy if y = x +1  x 3 - 1 dx 2 1Example 2: , ( x ≠ 1) x2 -1Solution: Given that( ) ( ) ( )==y  3 - 1 x +1 x 2 x +1  x 3 - (1)3  1 x -1 x2 -1 ( )( )( ) ( )( )= x +1 x -1 x +1+ x x +1 x +1+ x x -1 = = ( x +1)( )(x -1 x +1+ x=) ( ) (x +1 2 + )x +1 x 31 = x +1+ 2 x + x x + x= x2 + 2x + 2x2 +1 =dy d  3 + 2x + 1 +=1 d  x 3  + d (2x) + d  1  + d (1) dx dx dx 2 dx dx dx x2 2x2 2x2 = 3 1 + 2(1) + 2. 1 + =0 3 x +2+ 1 2 x2 2x 2 x 25

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab ( )Diferentiate x +1  x 3 - 1 eLearn.Punjab 2 version: 1.1 3Example 3: 1 with respect to x. x2 - x2( )Solution: Let y = x +1  3 - 1 x2 31 x2 - x2( )= 3 - 1 x +1 x 2 x ( x -1) ( x +1)( x -1)( x + x +1) (( x -1) x + x +1) x ( x -1) x ( x -1) = x + x +1 xDiferentiating with respect to x , we havedy = d  x + x + 1dx dx  x (x d x + x +1) - ( x + x + 1) d ( x) dx dx= ( )2 x( )= x 1 + 1 - 1 + 0  - x+ x +1 . 1 - 1  2 2 2 2 x x x( )= x 1 + 2 1  - x+ x +1 1 x 2x x 26

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab===x  2 2 x + 1  - x + x +1 eLearn.Punjab x 2 x version: 1.1 2x + x - x - x -1 x -1 x x.2 x 3 2x2Example 4: Diferentiate 2x3 - 3x2 + 5 with respect to x. x2 +1Solution: Let f ( x ) x=)2=x23 xx-233-+x132 x+25+. Then we take f ( 5 and g( x=) x 2 +1 ( )f ' ( x)= d 2x3 - + 5= - 3(2x) + 0= 6x2 - 6xNow dx 3x2 2 3x2and g'(x) = d  x2 + 1 = 2x + 0 = 2x dxUsing the quotient formula: f '(x) = f '( x ) g ( x) - f( x ) g ' ( x ) ,we obtain g ( x )2( )( ) ( )d( )dx 2 x3 - 3x2 + 5  = 6x2 - 6x x2 +1 - 2x3 + 3x2 + 5 (2x) x2 +1 x2 +1 2 ( )( )=6x4 - 6x3 + 6x2 - 6x - 4x4 - 6x3 + 10 x x2 +1 2 6x4 - 6x3 + 6x2 - 6x - 4x4 + 6x3 - 10 x x2 +1 2 ( )= + 6x2 -16x 2x4 x2 +1 2 ( )= EXERCISE 2.3 Diferentiate w.r.t. x1. x4 + 2x3 + x2 2. x-3 + 2x-3/2 + 3 3. a+x a-x 27

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab4. 2x -3 5. ( x - 5)(3 - x) 6.  x- 1 2 2x +1 x( )7.1+ x  x - x 3  ( )x2 +1 2 2 8. x2 -1 x 9. x2 +1 x2 - 310. 1 + x 11. 2x -1 12. a-x 1- x x2 +1 a+x13. x2 + 1 14. 1+ x - 1- x 15. x a+x x2 -1 1+ x + 1- x a-x16. If=y x - 1 , show that 2x dy + y =2 x x dx17. If y =x4 + 2x2 + 2 , prove tha=t dy 4x y -1 dx2.4 THE CHAIN RULE The composition fog of functions f and g is the function whose values f [g(x)], are found( )for each x in the domain of g for which g(x) is in the domain of f . f g ( x) is read as f of gof x).Theorem. If g is diferentiable at the point x and f is diferentiable at the point g( x ) thenthe composition function fog is diferentiable at the point x and ( fog )'( x) = f 'g ( x).g '( x) .The proof of the chain rule is beyond the scope of this book. =If y (=fog )( x) f g ( x) , then ==( fog )'( x)  f g ( x) ' dy dx ⇒ dy =f 'g ( x ).g '( x ) (i) dx (ii) Let u = g ( x) (iii) Then y = f (u) version: 1.1 28

21.. DQifuaerdernattiiactEioqnuations eLearn.PunjabDiferentiating (ii) and (iii) w.r.t x and u respectively, we have. eLearn.Punjab version: 1.1=du dd=x g ( x) g '(x) dx=and dy dd=u  f (u) f 'u duThus (i) can be written in the following forms(a) d ( f (u)) = f '(u ) du dx dx(b) dy = dy .du dx du dxThe proof of the Chain rule is beyond the scope of this book.Note=:=1. Let y g ( x)n and u g ( x) =Then y u=n and dy nun-1 (power rule) du Bu=t dy d=y . du nun-1 du dx du dx dx or d  g ( x ) n = n g ( x)n-1 .g' ( x)  du = g ' ( x)  dx dx 2. Reciprocal rule can be written as d  g 1 )  = d  g ( x )  -1= - 1.g ( x)-1-1 .g' ( x) dx dx (x = (-1) g ( x)-2 .g' ( x)Example 1: ( )Find the derivative of x3 +1 9 with respect toSolution: ( )Let y+=x3 1 +9 =and u x=3 1 Then y u9 29

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab ==Now du 3x2 and dy 9u8 (Power formula) dx duUsing the formula dy = 9u8 du , we have dx dx( ) ( ) ( )or u du =3x2  d x3 + 1 9 =9 x3 + 1 8 3x2 =x3 +1 and dx dx ( )= 27x2 x3 +1 8Example 2: Diferentiate a-x ,( x ≠ -a) with respect to x a+xSolution: Let y= =aa=+- xx and u a - x . Then y 1 a + x u2 No=w dy 1=u 12 - 1 1 - 1 2 u du 2 2 =and du dd=x  aa +- xx   d ( a - x ) ( a + x) - (a - x)  d (a + x )  dx dx dx (=0=-1)(a +(xa)+-x( ) ( 1) (a + x)2= a - x 0 + -a - x - a + x -2a )2 (a + x)2 (a + x)2Using the formula dy = dy . du , we have dx du dxd  a - x  =1 - 1  -2a  =1  a - x - 1 × -2a u =a - x dx a + x 2 2 a+x 2 a + x 2 a + x u a + x)2 ( )2 ( = (a - x)- 1 × -a = -a (a + x)- 2 (a + x)2 ( a - x )1 ( a + )3 1 2 2 x2 version: 1.1 30

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.PunjabExample 3: Find dy if y = a+ x + a-x (x ≠ 0) dx a+ x - a-xSolution: y= a+ x + a-x a+ x - a-xMultiplying the numerator and the denominator by a + x - a - x , gives (( ))(( ))y = a+x+ a-x a+x- a-x a+x- a-x a+x- a-x ( ) ( )= ( a + a+x -2 =-x=) -a2 -ax2 -2 x2 (a + x) -(a - x) 2x x) + (a 2a - 2 a2 - x2 ( )2 a - a2 - x2 that is, y = x a - a2 - x2Let f ( x) =x and g( x) =a - a2 - x2 , then( )f ( x)' =1 and -g'( x) =0- d- 1 d1 - 1 2 2 =a2 x-2 ( ) ( )1a2-x2 a2 x2 dx 2 dx -= 1 -x ( 2=x ) x 2 a2 - x2 a2 - x2Using the formula dy = f '( x) g(x)- f (x) g '(x), we have dx g ( x)2 1. a - a2 - x2 - x. x a2 - x2 ( )dy = ( )dx 2 a - a2 - x2 ( )= a a2 - x2 - a2 - x2 - x2 = a a2 - x2 - a2 2 ( )2 ( )a2 - x2 a - a2 - x2 a2 - x2 a - a2 - x2 version: 1.1 31

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab -a a - a2 - x2 = -a( )= 2 ( )2 ( )a2 - x2 a - a2 - x2 a2 - x2 a - a2 - x2Example 4: ( )Find dy if y= 1+ 2 33 dx x .x2( ) ( )Solution: + y =1 +2 =x 3 .x32  1  1   3 2 2 x x ( )1 (i)  ( ii ) Let u= 1 + 2 x .x2Then y = u3Differentiating (ii) with respect to u, we have( ) ( )dd+yx==3u2 3+ =1 2 x 1 2 31 2  x2 2 x .xDiferentiating (i) with respect to x , gives( )du =  0 + 2. 1  1 + 1+ 2 x 1 2xdx 2 x x2+ =1 1=+ 2 x 2 x +1=+ 2 x 1 + 4 x 2x 2x 2xUsing the formula dy = dy . du ,we have dx du dx  x 3 .x 32+ =3 1 2  + ( ) ( )d 1+ 2 x 2 x 1 2 4 x xdx .x ( ) ( )+ =3 1 2 x+2 x 1 4 x 2 (+=1 2 x+) ( x 4x)Example 5: If y = (ax + b)n where n is a negative integer, ind dy using quotient theorem dxSolution: Let n = -m where m is a positive integer. Then version: 1.1 32

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjaby =(ax + b)n = ( ax + )b -m =( ax 1 b)m (i) eLearn.Punjab + version: 1.1We first find d (ax + b)m . Let+u =ax b. Then dx( ) ( )d (ax + b=)m d um= d um du (using chain rule)dx dx dx dx = +mum-1 x a=m(ax )b m-1.a  d ( ax + b ) =a  dxNow diferentiating (i) w.r.t.’ x ’, we have==dy d  ( ax 1 b)m  d (1).(ax + b)m -1. d (ax + b)m dx dx + dx dx (ax + b)m 2 0.( ax + b)m -1.m(ax + )b m-1 .a (ax + b)2m( )= m ( ax- =m+(ax )b m-1 .a x+(ax- )b=-2m+ )b m-1-2m .a=(-m ) ( ax + b )-m-1 . a+=n( ax b )n-1 .a =(-m n )Example 6: Find dy if y = xn where n = p , q ≠ 0 dx qSolution: Given that y =xn wher≠e n =p , q =0. putting n p ,we have q q p (i) y = xqTaking qth power of both sides of (i), we get yq = xp (ii)Diferentiating both sides of (ii) w.r.t. ‘ x ‘ , gives( )d (yq ) = d (x p ) or d (yq ) . dy = d x p (Using chain rule)dx dx dy dx dx⇒ q yq-1 dy = px p-1 (iii) dx 33

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.PunjabMultiplying both sides of (iii) by y, we have q . yq dy = py x p-1 or q. x p dy = p. x x p-1 (using (i) and (ii)) dx dx ⇒ dy = p. 1 p =p × x p +p-1-p dx q xp q q . x q x p-1 p p -1 nx n-1  p = n q q q = x = Thus d (xn ) n xn-1 . dx2.5 DERIVATIVES OF INVERSE FUNCTIONS If for each x d Df , f(x) = y and for each y d Dg, g(x) = x, then f and g are inverse of eachother, that is, ( g o f=)( x ) g(=f ( x )) g=( y ) x (i)and ( f o=g)(y) f (g=(y)) f=(x) y (ii)Using chain rule, we can prove that f '( x ). g'( y ) = 1⇒ f ' (x) = 1 g' (y)⇒ dy = 1 and f (x) = y ⇒ f ' (x) = dy  dx dx g(y) = x ⇒ g'(y) = dx dy dx dy2.6 DERIVATIVE OF A FUNCTION GIVEN IN THE FORM OF PARAMETRIC EQUATIONS The equations x = at2 and y = 2at express x and y as function of t . Here the variable tis called a parameter and the equations of x and y in terms of t are called the parametricequations. version: 1.1 34

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab Now we explain the method of inding derivatives of functions given in the form ofparametric equations by the following examples.Example 1: Find dy if x = at2 and y = 2at. dxSolution: We use the chain rule to find dy dx Here dy = d (2at) = 2a.1=2a dt dt and dx = d (at2 ) = a (2t) = 2at dt dt dyso dy = dy . dt = dt = 2a = 2a (2a = y) dx dt dx dx 2at y dtEliminating t, we get x = a  y 2 = a. y2 = y2 ⇒ y2 = 4ax (i) 2a 4a2 4aDiferentiating both sides of (i) w.r.t. ‘ x ’ we haved (y2 ) = d (4ax) ⇒ 2y dy = 4a (1)dx dx dxd (y2 ) . dy = 4a d (x)dx dx dx ⇒ dy = 2a dx yExample 2: Find dy if x 1 - t2 and y = 3t2 - 2t3 . dxSolution: Given that x = 1 - t2 ...... (i) and y = 3t2 - 2t2 (ii)Diferentiating (i) w.r.t. ‘t ’ ,we get version: 1.1 35

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab( ) ( )dy =d 1- t2 =d (1)- d t2 =0- 2t =- 2t eLearn.Punjab dt dt dt dt version: 1.1Diferentiating (ii) w.r.t. ‘t ’ ,we have( ) ( ) ( )dy= d 3t2 - 2t2 = d 3t2 - d 2t3 dt dt dt dt( )=3(2t )- 2 3t2 =6t - 6t2 =6t (1- t )Applying the formula dy=dy d=y . dt dt dx dt dx dx dt=6t-(12-t t ) =-3(1-t ) =3(t -1)Example 3: Find=dy if x 11=+-tt22 , y 2t dx 1+ t( )==Solution: Given that x 1+ t 2 (i) and y 2t (ii) 1+ t 2 1+ t 2Diferentiating (i) w.r.t. ‘t ’ ,we get( ) ( ) ( ) ( )=dx  d 1- t 2  1+ t2 - 1- t 2 .d 1+ t2 dt dt dtdd=t  11+-tt22  (1 + t2 )2==(-2t )(1 )+ t2 - (1- t 2 ) ( 2t ) ( )2t -1- t2 -1+ t2 -4t (1 + t2 )2 ( )1+ t2 2 ( )1+ t2 2Diferentiating (i) w.r.t. ‘t ’ ,we have 36

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab  d ( 2t )  (1 + t 2 ) - 2t x d (1 + t 2 ) dt dt==dy d  2t  ( )1+ t2 2 dt dt 1 + t2( ( ) ) ( )= 2(1- t2 ) 2=1 +1t=2+ t-2 22t ( 2t )=2 + 2t2 - 4t 2 2 - 2t2 ( )1+ t2 2 1+ t 2 2 ( )1+ t2 2 dy 2(1- t2 )( ) ( )dy= dd=yt 2( )dx dy . =dt dx 1 + t2 2 1--4tt=2 t2 -1 dt dx 4t = 2t - 1+ t2 22.7 Differentiation of Implicit Relations Sometimes the functional relation is not explicitly expressed in the form y = f ( x)but an equation involving x and y is given. To ind dy from such an equation, we diferentiate dxeach term of the equation and use the chain rule where it is required.The process of indingdy in this way, is called implicit diferentiation. We explain the implicit diferentiation in thedxfollowing examples.Example 1: Find dy if x2 + y2 =4 dxSolution: Here x2 + y2 =4 (i)Diferentiating both sides of (i) w.r.t. ‘ x ‘ , we get version: 1.1 37

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab 2x + 2 y dy =0 dxor x + y dy =0 ⇒- dy =x dx dx ySolving (i) for y in terms of x, we have (ii)y±=-4 x2 (iii)⇒ y = 4 - x2or y-=-4 x2dy found above represents the derivative of each of functions deined as in dxdx(ii) and (iii)From (ii) dy = 1 x ( --2 x ) = x x2 dx 2 4 - x2 - 4( )=- x  4 - x2 =y y -x =- x 4 - x2 y( )From (iii) dy =- 1 dx 2 4 - x2 x (-2x) =- - 4 - x =yExample 2: Find dy ,if y2 + x2 - 4x =5. dxSolution: Given that y2 + x2 - 4x =5 (i)Diferentiating both sides of (i) w.r.t. ‘ x ’ ,we get d  y2 + x2 - 4x =d (5) dx dx ( ) ( )=ddx y2 dy or 2 y dy + 2x - 4=0 d=y2 dy 2 y dx  dx dx dx version: 1.1 38

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab⇒ 2 y dy =4 - 2x ⇒ dy =2(2 - x) =2 - x (ii) eLearn.Punjab dx dx 2 y y version: 1.1Note: Solving (i) for y , we havey2 =5 + 4x - x ⇒ y =± 5 + 4x - x2Thus y = 5 + 4x - x2 (iii)or y =- 5 + 4x - x2 (iv)Each of these equations (iii) and (iv) deines a function.Let y = f1 ( x) = 5 + 4x - x2 (v) (vi)and y =f1 ( x) =- 5 + 4x - x2 .Diferentiation (v) w.r.t. ‘ x ‘ , we get ( )f1' ( x)= 1 - 1 × (4 - 2x)= 2-x 5 + 4x - x2 2 5 + 4x - x2 2From=(v)=, 5 + 4x - x2 y, so f1' ( x) 2-x y( )Also ( ) 1 - 1 ( - 2x) =- 2-x f2 ' x =- 2 5 + 4x - x2 2 × 4 5 + 4x - x2From (vi) -=5 +=4x - x2 y, so f2' (x) 2-x yThus (ii) represents the derivative of f1 ( x) as well as that of f2 ( x).Example 3: Find dy if y2 - xy - x2 + 4 =0. dxSolution: Given that y2 - xy - x2 + 4 = 0 (i)Diferentiating both sides of (i) w.r.t. ‘ x ‘ , gives 39

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab d  y2 - xy - =x2 + 4 d (0=) 0 dx dxor 2y dy - 1.y + x dy  - 2x + 0 =0 dx dx⇒ ( 2 y - x) dy =+2x y ⇒ dy =22xy +- y dx dx xExample 4: Find dy if y3 - 2xy2 - x2 y + 3x =0. dxSolution: Diferentiating both sides of the given equation w.r.t. ‘x’ we have d  y3 - 2 xy 2 + x2 y + 3=x d (0)= 0 dx dxor d ( y3 ) - d ( 2 xy 2 ) + d ( x2 y ) + d (3x) =0 dx dx dx dx ( ) ( )d y3 - 2 1.y 2 + x d y2  +  2xy + x2 dy  + 3 =0 dx  dx dxUsing the chain rule on d y3 and d y2 , we have ( ) ( )dx dx 3y2 dy - 2  y2 + x  2 y dy  + 2xy + x2 dy + 3 =0 dx dx dx( )or 3y2 - 4xy + x2 dy = 2 y2 - 2xy - 3 dx⇒ dy =32yy22--42xxyy+-x32 dxExample 5: Diferentiate x2 + 1 w.r.t. x - 1 x2 xSolution: L+et y =x-2 x12=and u x 1. Then x version: 1.1 40

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab ( ) ( )( )dy = dx 2x + ( -2 ) . 1 = 2 x - 1  = 2 x4 -1 = 2 x2 -1 x2 +1 x3 x3 x3 x3 and du = 1 - ( -1) . 1 = 1+ 1 = x2 +1 dx x2 x2 x2 ( )( ) ( )Thus d=y du 2 x2 -1 x2 =+ 1 x2 =2 x2 --1 2 x  dy . d=x x3 . x2 +1 x 1 dx du x EXERCISE 2.41. Find dy by making suitable substitutions in the following functions deined as: dx (i) y= 1- x (ii) =y x + x (iii) y=x a+x 1+ x a-x ( )(iv) y= 3x2 - 2x + 7 6 (v) a2 + x2 a2 - x2. Find dy if: dx (i) 3x + 4 y + 7 =0 (ii) xy + y2 =2 (iii) x2 - 4xy - 5y =0 (iv) 4x2 + 2hxy + by2 + 2gx + 2 fy + c =0 ( )(v) x 1+ y + y 1+ x =0 (vi) y x2 -1= x x2 + 43. Find dy of the following parametric functions dx ( )(i) x= q + 1 and y = q +1 ==(ii) x a 1-t2 ,y 2bt q 1+ t2 1+ t24. Prove that y d=y + x 0 =if x 1 - t 2 =, y 2t dx 1 + t 2 1+ t version: 1.1 41

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab5. Diferentiate(i) x2 - 1 w.r.t x4 ( )(ii) 1 + x2 n w.r .t x2 x2(iii) x2 +1 w.r .t x -1 (iv) ax + b w.r .t ax2 + b x2 -1 x +1 cx + d ax2 + d(v) x2 +1 w.r .t x3 x2 -12.8 DERIVATIVES OF TRIGONOMETRIC FUNCTIONSWhile inding derivatives of trigonometric functions, we assume that x is measured inradians. The limit th=e=orems lim sin x 1 and lim1 - cos x 0 are used to ind the derivativeformulas for sin x and cos x. x→0 x x→0 xWe prove from irst principle that d ( sin x) = cos x and d (cox x)= - sin x dx dxLet y = sin x Then y+ d y = sin ( x + d x)and d y= sin( x + d x) - sin x= 2 cos  x + dx + x =sin x + dx - x  +2cos  x dx  sin  dx  2 2 2 2 =dd xy 2 cos  x + dd=2xx  sin dx  +cos  x dx  sin  dx  2 2 d 2 x 2 =dlxi→m0 dd xy lim   x + dx  sin  dx   cos 2 d 2  d x→0 x  2  version: 1.1 42

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab = +dlxim→0 cos  x dx  lim sin  dx  whd2exnd→x 0  2 2 d 2 → d x →0 x 0 22=Thus =ddyx co=s x+.1.d xli/ 2m→0 cos  x dx  sin d x 1 2 2 cos x and lim d x d x / 2→0 2Let y= cos x, then y + d y= cos( x + d x)and d y = cos( x + d x)- cos x = cos x cosd x - sin x sind x - cos x- =s-in x sind x cos x  1 - cos d x  dx d y =( sin x). sidn xd-x cos x 1- cdo-xsd x  d xlim d y =d-lxi→m0 ( sin x) s-idndx x cos x  1 - cos d x  d x dxd x→0 = lim ( - sin x ) sin d x  - lim - cos x  1 - cos d x  d x  dx d x→0 d x→0Thus d-y =( si-n x).1 (cos x)(0) dlxi→mdlxi0→m01s-indcdoxdxxyd=x1a=nd0  dxor d (cos x) = - sin x dxNow using d ( sin x ) = cos x and d (cos x)= - sin x, we prove that dx dx==d (sec x) sec x tan x and d (cot x) cosec2 x dx dx version: 1.1 43

12.. DQiufaedrernattiiactEioqnuations eLearn.PunjabProof of d ( sec x) = sec x tan x. eLearn.Punjab dx version: 1.1Let =y s=ec x 1 (i) cos xDiferentiating (i) w.r.t. ‘ x ’ , we have=ddx ( y) dd=x  co1s x   d (1) cos x -1. d (cos x)  Using   dx dx  quotient  = 0.cos x -1.(- formula sin x) (cos x)2 cos2 x== 1 . sin x sec x tan x cos x cos xThus d ( sec x) = sec x tan x dxProof of d (cot x) = cos ec2 x dxLe=t y c=ot x cos x (i) sin xDiferentiating (i) w.r.t. ‘ x ’ , we get=d ( y) dd=x  csoins xx   d (cos x ) sin x - cos x d ( sin x)  Using  dx dx quotient dx (sin x)2 formula = (- sin x) sin x - cos x (cos x) sin2 x ( )=- sin2six-n+2 c-os2 x==sin12 x cos ec2 x xThus d (cot x) = cos ec2 x dx 44

21.. DQifuaerdernattiiactEioqnuations eLearn.PunjabNow we write the derivatives of six trigonometric functions eLearn.Punjab version: 1.1(1) d (sin x) = cos x (2) d (cos x) = sin x dx dx(3) d (tan x) = sec2 x (4) d (cot x) = - cosec2 x dx dx(5) d (cosec x) = - cosec x cot x (6) d (sec x) = sec x tan x dx dxExample 1: Find the derivative of tan x from irst principle.Solution: Let y= tan x, then +y =d x tan +( x d x) and d y = y + d x - y = tan ( x + d x) - tan x = sin( x + d x) - =sin x sin( x + d x)cos x - cos( x + d x) sin x cos( x + d x) cos x cos( x + d x) cos x== sin( x + d x - x) sind x cos( x + d x).cos x cos( x + d x) cos x d y = 1 . sind x d x d dx cos ( x + x ) .cos xor lim d y = lim  1  . lim  sin d x  d x d x d x→0 d x→0 cos( x + d x).cos x d x→0==Thus dy 1  lim cos ( x + d x) = cos x  dx x 1 x)(cos x) .1 sec2 x d x→0 d = x (cos  and lim sin d d x→0Thus dy = sec2 x or d (tan x) = sec2 x dx dx 45

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.PunjabExample 2: Diferientiate ab-initio w.r.t. ‘ x ‘(i) cos 2x (ii) sin x (iii) cot 2 xSolution: (i) Let=y cos 2 x, then y+ d=y cos 2( x + d x) and d y= cos(2x + 2d x) - cos 2x- =2 sin 2x + 2d x + 2x sin 2x + 2d x-- 2x =2+sin(2x d x) sind x 22Now- dy =2+sin ( 2 x d x ). sind x dx dxThus dy =lim-  2 sin(2+x d x ) . sind x  dx d x→0 dx- =2+dlxi→m0( sin 2x d x ) . lim sind x dx d x→0-=( 2 sin-2x=).1 2 sin 2 x  lim sin ( 2 x + d x) =sin 2x and lim sind x 1 = d x→0 dx d x→0(ii) Let y +sin x , then y + d y = sin x d xand d =y sin x + d x - sin x = 2cos  x+dx + x  sin x+dx - x  2 2( )( )As x + d x + x x + d x - x =( x + d x) - x =d x,So d y = 2 cos  x+dx + x . sin x+dx - x  d x 2 2 x  dx ( )( )= 2cos x+dx + x  sin x+dx - 2 2 x+dx + x x+dx - x version: 1.1 46

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab cos  x + dx + x  sin  x + dx - x  + d 2 + d 2 x x . x x x+ x- 2  x+dx + x  . lim  sin  x +dx - x    2  0 + 2  Thus dy →lim cos x+dx - x dx +dx + 2 dx - d x→0 x x x 2 x  cos x+ x   x + dx - x →    .1 0==dy 2 cos x 2 0 when dx + 2x dx→ x x(iii) Let y = cot2x, theny + d=y cot2 ( x + d x) d =y cot2 ( x + d x) - cot2=x cot ( x + d x) + cot x x c+ot ( x - d x) cot x = cot (x +d x) + cot x . cos( x ++dd-xx)) cos x  sin ( x sin x = cot ( x + d x) + cot x × sin x cos( x + d x) - cos x sin( x + d x) sin( x + d x) sin xd y =  cot ( x + d x) + cot x  . - sind x  sin x cos( x+ d x) - cos x sin( x +d x) x d x sin( x + d x) sin x dx x-(x +d x =sin ( )) =sin(-d x) =- sind lim d y -lim  cot ( x + d x) + cot x .( 1) sind x  d x sin ( x + d x dx d x→0 d x→0 ) sin x =Thus dy -cot x + cot x .( 1). 1 andldxi→mdl0xi→mco0 tsi(nx( + d x ) =cot x  dx sin x sin x x + d x) =sin x = -2cot x .1 = -2cot x co sec2 x sin2 x version: 1.1 47

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.PunjabExample 3: Differentiate sin3 x w.r.t. cos2 xSoluti=o=n: Let y sin3 x and u cos2 xNow=dy 3sin2 x co=s x and du -2cos x( sin x) dx dx( )Thus=dy d=y . dx  dx 1  du dx du du dx 3sin2 x cos x .=-2cos1x sin x du = - 3 sin x. 22.9 DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONSHere we want to prove that1. d sin-1 x = 1, x ∈(-1,1) or -1 < x < 1 dx 1- x2 x ∈(-1,1) or -1 < x < 12. d Cos -1 x  = - 1, x ∈R dx 1- x2 x ∈[-1,1]' , [-1,1]' = (-∞,-1) ∪ (1,∞)3. d Tan-1x  = - 1 , x ∈[-1,1]' , [-1,1]' = (-∞,-1) ∪ (1,∞) dx 1+ x2 x ∈R4. d Cosec-1 x  = - 1, dx |x| x2 -15. d  Sec-1x  = - 1, dx |x| x2 -16. d Cot -1 x  = - 1 , dx + x2 1Proof of (1). Let y = Sin-1x (i). version: 1.1 48

21.. DQifuaerdernattiiactEioqnuations eLearn.PunjabT=he=n x Sin ∈y o-r x sin y for y  p ,p  (ii) eLearn.Punjab 2 2 version: 1.1Diferentiating both sides of (ii) w.r.t. ‘ x ’ , we get=1 dd=x (sin y) d (=sin y) ddyx cos y dy dx dx=⇒ dy ∈1- for y  p ,p  dx cos y 2 2 =1  cos y is positive for y ∈  - p ,p  1 - sin2 y 2 2( )Thus =d sin-1 x - 1< < for 1 x 1 dx 1 - x2Proof of (2). Let y = Cos -1x (i) (ii) =T=hen x Co∈s y or x cos y for y [0, p ]Diferentiating both sides of (ii) w.r.t. ‘ x ’ , gives1 = d (cos y) = d (cos y) dy= - sin y dy dx dx dx dx⇒ -dy =1 for y ∈(0,p ) dx sin y =- 1  sin y is positive for y ∈(0,p ) 1 - cos2 y for -1 <x <1( )Thus d Cos-1x =- 1 dx 1 - x2Proof of (3). Let y = Tan-1x (i).Then x==Tan y or x ta∈n y- for y  p ,p  (ii) 2 2Diferentiating both sides of (ii) w.r.t. ‘ x ’ , we have 49

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab=1 d=(tan y) d (=tan y) dy sec2 y dy eLearn.Punjab dx dx dx dx version: 1.1=⇒ dy s∈ec12-y for y  p ,p  dx 2 2= 1=+ ta1n2 y 1 for x ∈ R 1+ x2Thus =ddx Tan-1x ∈ 1 2 for x R 1 +xProof of (4). Let y = Co sec-1 x (i) (ii)The=n =x Co se∈c y -or x c-os ec y for y  p ,p  {0} 2 2 - p ,p  - {0} is also written as - p 0 ∪ 0,p2  2 2 2Diferentiating both sides of (ii) w.r.t. ‘ x ’ , we get==1 d (cosec y) d (cosec y) dy dx dx dx = ( - cosec y cot y) dy dx⇒ dy = - 1 for y ∈ -p ,p  -{0} 2 2dx cosec y cot yWhen y ∈  0,p  , cos ec y and cot y are positive. 2As cosec y = x , so x is positive in this caseand cot y= co sec2 y -1= x2 -1 for all x > 1( )=Thus d Co sec-1 x > -1 for x 1 dx x x2 -1 50

21.. DQifuaerdernattiiactEioqnuations eLearn.PunjabWhen y ∈  - p ,0  ,cosec y and cot y are negative eLearn.Punjab 2 version: 1.1As cosec y = x, so x is negative in this caseand cot y = - cosec2 y -1 = - x2 -1 when x < -1 < - -1 x - x2 -1( )Thu=s ddx Co sec-1 x (x 1) = <( --x ) -1 - 1 (x 1) x2d cosec -1 x  = - 1 for x ∈[-1,1]'dx |x| x2 -1Proof of (5). is left as an exerciseProof of (6). is similar to that of (4)Example 1: Find dy if =y x Sin -1 x  + a2 + x2 dx aSolution: Given that=y x Sin -1  x  + a2 + x2 aDiferentiating w.r.t. x , we have  x a=2 + x2   x ( )=dy d Sin-1 x + d Sin-1 x + d 1/ 2 dx dx a dx a dx a2 + x2 = 1 . Sin-1 x + x. 1 .d +ax  (1 .+a2 ) (x2 1 -1 d +a2 )x2 a 1 -  dx 2 2 x 2 dx a 51

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.PunjabSin-1 x + x 1 .1 + 1 .( -2 x ) a x2 a 2 a2 - x2 1- a2Sin-1 x + x a . 1 - 1 =Sin-1 x a a2 - x2 a a2 - x2 a==Example 2: If y tan  2 Tan-1 x  ,show that dy 4(1+ y2 ) 2 dx 4 + x2Solution: Let u = 2 Tan-1 x , then 2y =tan u ⇒ dy =sec2 u =1+ tan2 u =1 + y2 duan=d ddux =ddx  2 Ta=n-1 2x  2.1=+ 12x 2 . d  x  2 .1 4 dx 2 1+ x2 2 4 + x2 4( ) ( )Thus+dy ==dy . du=1 y2 . 4 4 x 2 4 1+ y2 dx du dx + 4 + x2 EXERCISE 2.51. Diferentiate the following trigonometric functions from the irst principle,(i) sin x (ii) tan3x (iii) sin 2x + cos 2x (iv) cos x2(v) tan2 x (vi) tan x (vii) cos x2. Diferentiate the following w.r.t. the variable involved(i) x2 sec 4x (ii) tan3q sec2 q(iii) (sin 2q - cos3q )2 (iv) cos x + sin x version: 1.1 52

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab3. Find dy if (ii) x = y sin y eLearn.Punjab dx version: 1.1 (i) y = x cos y4. Find the derivative w.r.t. x (i) cos 1+ x (ii) sin 1+ 2x 1+ 2x 1+ x5. Diferentiate (ii) sin2 x w.r.t. cos4 x (i) sin x w.r.t. cot x6. If tan y (1 + tanx-) =1 tan x,show that -dy =1 dx7. If=y tan x + tan x + tan x + ...∞,prove that-(2 y=1) dy sec2 x. dx8. If =x a cos3q =, y b sin3q , show t+hat a=dy btanq 0 dx9. Find dy if x =a(cos t + sin t ) , y =a(sin t - t cos t ) dx10. Diferentiate w.r.t. x (i) Cos-1 x (ii) Cot-1 x (iii) 1 Sin-1 a a a ax (iv) Sin-1 1 - x2 (v) Sec-1  x2 +1  (vi) Cot -1  2x  x2 -1 - x2 1 (vii) Cos-1  1 - x2  1 + x211=. dy y=if y Tan-1 x dx x x y( ) ( ) ( )12. If y t=an +p Tan-1-x , sho+w that 1 x2 y1 p 1 y2 0 53

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab2.10 DERIVATIVE OF EXPONENTIAL FUNCTIONS: A function f deined by f (x) = ax a > 0 , a ≠ 1 and x is any real number.is called an exponential function If a = e , then y = ax becomes y = ex . ex is called the natural exponential function. Now we ind derivatives of ex and ax from the irst principle:1. Let y = ex then y + d y =ex+d x and d y =y + d y - y =ex+d x - ex =ex . ed x - ex( )That is,d y-=ex ed x 1 an=d dd xy ex .  ed x - 1  d x==Thus dlxi→m0 dd xy lim e x  ed x - 1  ex . lim  ed x - 1  d x d x→0 d x d x→0d lim 0 ex = ex  x→==dy e x .1  Using lim eh -1 1 dx h h→0( )or d ex = ex dx2. Let y = ax , then y + d y= ax+d x and d y= a x+d x - a x= ax . ad x - ax= ( )ax ad x -1Dividing both sides by d x , we have dy = a x  ad x -1  dx d x==Thus ddyx =dlxi→m0 ax  addx x-1  ax . lim  ad x -1  d lim 0 ax ax  d x x→ d x→0 version: 1.1 54

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab = ax .(ln a )  Using lim=ah -=1 log a ln a  eLearn.Punjab h→0 h version: 1.1 e( )or d ax = ax .(ln a) dxExample 1: Find dy if : (i) y = ex2+1 (ii) y = a x dxSolution: (i) Let =u x2 +1 , then ( )y = eu ....(A) and d=u d x2 +=1 2x dx dxDiferentiating both sides of (A) w.r.t. 'x', we have=ddx ( y) dd=x (eu ) d ( eu ) . du (Using the chain rule) du dx( )= eu .du  d ex ex  dx dx UsingThus dy (=ex2+1 . 2x) + u x2 1 =a=nd du 2x  dx dx==(ii) Let u x Then y au ( A)( )and=du d =x1/ 2 1=x-1/ 2 1 dx dx 2 2xDiferentiating both sides of (A) w.r.t. 'x', gives=dy dd=x (au ) ddu=(au ) ddux  dy dy . du  dx dx du dx (au ) .du ( ) d ax ax ln a  ln a dx Using dx( ) ( )Thus=d a x =u 2 1 x  dx a x ln a =. 1 x and du 2x dx 55

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab = ln a . a x . 1 2xExample 2: Differentiate y = ax w.r.t. x.Solution: Here y = ax = ex ln aDiferentiating w.r.t. ‘ x ‘ , we have dy = ex ln a , d ( x ln a ) dx dx (ex In a )a x (ex In a )a x== ax .(In a)== ax .(In a)2.11 DERIVATIVE OF THE LOGARITHMIC FUNCTIONLogarithmic Function:If a > 0 a ≠ 1 and x =a , then the function defind by=y >logax (x 0)is called the logarithm of x to the base a.The logarithmic functions log e x and log x are called natural and common logarithms 10respectively, y =logex is written as y = ln x .We first find d ( In x). dxLet y = ln x Then and y + d y= In ( x + d x) dy= ln ( x + d x) - ln x =  x +d x  = ln 1 + dx  x x version: 1.1 56