diff cal

MATH 401 – DIFFERENTIAL CALCULUS Definition2.2. The derivative of y = f(x) at point P on the curve is equal to the slope of the tangent line at P, thus the derivative of the function f given by y= f(x) with respect to x at any x in its domain is defined as: f 'x  dy  lim limy  f x  x  f x x0 dx x x0 x provided the limit exists. Example1. Differentiate the function y  3x Solution: y  3x y  y  3x  x y  y  3x  3x y  3x  3x  y Since y  3x then y  3x  3x  3x y  3x y  3 x dy  lim y  lim 3  3 dx x0 x x0 dy  3 dx Example2. Differentiate the function y  x2  x Since y  x2  x then Solution: y  x2  x y  y  x  x2  x  x y  y  x2  2xx x2  x  x y  x2  2xx x2  x  x  y y  x2  2xx x2  x  x  x2  x y  2xx x2  x y  x2x  x 1 y  2x  x 1 x 48

MATH 401 – DIFFERENTIAL CALCULUS dy  lim y  lim 2x  x 1  2x  0 1  2x 1 dy x0 x x0 dy  2x 1 dy 2.2 The Increment Method Based on the definition of the derivative, y' f 'x  lim f x  x f x x0 x we can write the steps in solving for this derivative. These are as follows: Step 1. Write down the expression for f (x  x)  f (x) and simplify. Step 2. Divide the result in Step 1 by x , again simplify. Step 3. Find the limit of the result in Step 2 as x approaches zero. The obtained limit is the derivative. Example1. Given find the first derivative f '(x) using the three- Solution: step rule. Step1. Step2. Step3. lim f x  x  f x  lim 2x  5  x x0 x x0 lim f x  x  f x  2x  5  0 x0 x lim f x  x  f x  2x  5 x0 x f 'x  lim f x  x  f x x0 x f 'x  2x  5 49

MATH 401 – DIFFERENTIAL CALCULUS Example2. Given y  1 , find y ' . x 1 Solution: Step1. f x  x f x  1  1  x 1  x  x 1 x  x 1 x 1 ( x  x 1)( x 1)  x 1  x  x 1  x 1  x  x 1 ( x  x 1)( x 1) x 1  x  x 1    2 2  x 1  x  x 1 ( x  x 1)( x 1)[ x 1  x  x 1]  (x 1)  (x  x 1) ( x  x 1)( x 1)[ x 1  x  x 1]  x 1 x  x 1 ( x  x 1)( x 1)[ x 1  x  x 1] f x  x f x   x ( x  x 1)( x 1)[ x 1  x  x 1] Step2. f x  x f x   x  1 x ( x  x 1)( x 1)[ x 1  x  x 1] x f x  x f x  1 x ( x  x 1)( x 1)[ x 1  x  x 1] Step3. lim f x  x  f x  lim 1 x0 x (x0 x  x 1)( x 1)[ x 1  x  x 1]  1 ( x  0 1)( x 1)[ x 1  x  0 1]  1 ( x 1)( x 1)[ x 1  x 1]  1  1  1 (x 1)[2 x 1] 2(x 1)(x 1)1 2 2(x 1)3 2 lim f x  x  f x  1 x0 x 2(x 1)3 2 y' lim f x  x  f x x0 x y' 1 2 2(x 1)3 50

MATH 401 – DIFFERENTIAL CALCULUS Exercise1. Find the derivative of the following using the definition of derivative. 1. y  1 x2 Answer: -2x 2. f x  x2  x Answer: 2x +1 3. y  x Answer: 1 2x 4. f x  1 Answer: 1 3x 3x2 5. y  x Answer: x  2 x 1 2x 13 2 51

MATH 401 – DIFFERENTIAL CALCULUS 2.3 Theories on Differentiation of Algebraic Functions The increment-method (three-step rule) of finding the derivative of a function gives us the basic procedures of differentiation. However these rules are laborious and tedious when the functions to be differentiated are “complex”, that is, functions with large exponents, functions with fractional exponents and other rational functions Understanding of the theorems of differentiation is very important. This is the heart of differential calculus. All of the succeeding topics such as applications of derivatives, differentiation of transcendental functions etc. will be dependent on these theorems. Understanding of these theorems will enable us to calculate derivatives more efficiently and will make calculus easy and enjoyable. Here we will be using the “dy/dx” notation (also called Leibniz's notation) instead of limits. The following are the rules of differentiation for Algebraic Functions. Theorem 2.1. The Constant Rule of Differentiation If c is a constant and any real number, then d (c)  0 dx Illustration 2.1. c.) d 3   0 a.) d 5  0 dx dx  d.) d 7  0 b.) d   3   0 dx dx  4  Theorem 2.2. The Power Rule (for positive integer powers) of Differentiation  d xn  nxn1 If n is a positive integer, then dx In words, to differentiate a power function, decrease the constant exponent by one and multiply the resulting power function by the original exponent. Illustration 2.2.  d a.) dx x2  2x21  2x 52

MATH 401 – DIFFERENTIAL CALCULUS  b.) d x4  4x41  4x5 dx c.) d  x 6   6 6 1  6 67  6 1 dx 7 x7 x7 x7 7 7 7 d.) d  x 3    3 x 3 1   3 x 32   3 x 5 dx 2 2 2 2 2 2 2 Theorem 2.3. The Constant Multiple Rule of Differentiation If f is a differentiable function at x and c is any real number, then cf is also differentiable at x and d cf ( x)  c d  f (x) dx dx In words, the derivative of a constant times a function is the constant times the derivative of the function, if this derivative exists. Illustration 2.3.    a) d 5x8  5 8x7  40x7 dx    b) d  9x4  9  4x5  36x5 dx c) d   5x 2   5 2 x 3   2x 3 dx 5 5 5 5  d) d  4 r3   4  3r 2  4r 2 dx  3  3 Theorem 2.4. The Sum Rule of Differentiation If f and g are both differentiable functions at x, then so are f + g and f – g, and d  f  g  d  f  d g or dx dx dx d  f (x)  g(x)  d  f (x)  d g(x) dx dx dx In words, the derivative of a sum or of a difference equals the sum or difference of their derivatives, if these derivatives exist. Illustration 2.4. 53

MATH 401 – DIFFERENTIAL CALCULUS a) d 3x  4  d 3x  d 4  3  0  3 dx dx dx dy b) d d d 3 d dx   2x 2 3  9  dx  2x2  dx 4x  dx  5x 2   dx 9  4x 5x 2  22 x   41  5 3 x 1   0 2 2  dy   2x2  4x  3  9  4x  4  15 x 1 dx 2 2 5x 2  c) d r2  4 r3   d r 2  d  4 r3  dx  3  dx dx  3     2r3  4  3r2  3 d r 2  4 r3   2r 3  4r 2 dx  3  Theorem 2.5. The Product Rule of Differentiation If f and g are both differentiable functions at x, then so is the product f  g , and d  f  g   f  dg  g  df or dx dx dx d  f (x)  g(x)  f (x)  d [g(x)]  g(x)  d  f (x) dx dx dx In words, the derivative of a product of two functions is the first function times the derivative of the second plus the second function times the derivative of the first, if these derivatives exist. Illustration 2.5. a) d 3x  44x2  3  3x  4 d 4x2  3 4x2  3 d 3x  4 dx dx dx  3x  48x  0 4x2  33  0  3x  48x 4x2  33  24x2  32x 12x2  9  d 3x  4 4x2  3  36x2  32x  9 dx 54

MATH 401 – DIFFERENTIAL CALCULUS b) d x3 15 - 2x  x3 1 d 5 - 2x  5 - 2x d x3 1 dx dx dx  x3 1 0 - 2 5 - 2x 3x2  0  x3 1 - 2 5 - 2x 3x2   2x3  2 15x2  6x3  d x3 1 5 - 2x  8x3 15x2  2 dx Theorem 2.6. The Quotient Rule of Differentiation If f and g are both differentiable functions at x, and if g(x) ≠ 0 then f g is differentiable at x and d  f   g  df  f  dg or dx g dx dx g2 d  f (x)  g(x)  d f (x) f (x)  d g(x) dx  g(x)  dx g ( x)2  dx In words, the derivative of a quotient of two functions is the fraction whose numerator is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator and whose denominator is the square of the given denominator. Illustration 2.6. 1 2x d  d 1 2x dx dx    a) d  4x2 3  4x2  3  4x2 3 dx 1 2x  1 2x2  1  2x8x  0  4x2  30  2 1 2x2  1  2x8x  4x2  3 2 1 2x2  8x  16x2   8x2  6 1  2x2  8x 16x2  8x2  6 1 2x2 55

MATH 401 – DIFFERENTIAL CALCULUS d  4x2 3    8x2  8x  6 dx 1 2x 1 2x2 d 3x2  4x  3x2  4x  d 3x  2 dx dx 3x  22    b)d  3x2  4x  3x  2 dx 3x  2   3x  26x  4  3x2  4x3 3x  22     9x2 18x212x 12x  8   12 x 3x  22 18x 2 8  9x2  12 x  3x  22 d  3x2  4x   9x2  12x  8 dx 3x  2 3x  22 Theorem 2.7. The General Power Rule of Differentiation If n is a positive integer and f is a differentiable functions at x, then d  f xn  n  f x  n1  d f x dx dx In words, the derivative of the power of a function is equal to the power, times the function raised to the power minus one, times the derivative of the function. Illustration 2.7.      a) d 2x3  4x2  2x 5  5 2x3  4x2  2x 51  d 2x3  4x2  2x dx dx     5 2x3  4x2  2x 4 6x2  8x  2     d 2x3  4x2  2x 5  5 6x2  8x  2 2x3  4x2  2x 4 dx b) d 3 5x  4  d 5x  1  1 5x  4 11  d 5x  4 3 43 dx dx 3 dx  1 5x  2  5  0 3 43 56

MATH 401 – DIFFERENTIAL CALCULUS d 3 5x  4  5 5x  2 dx 3 43 Theorem 2.8. The Derivative of a Composite Function If the function g is differentiable at x and the function f is differentiable at g(x), then the composition function f  g is differentiable at x and  f  g'x  f 'gxg'x Illustration 2.8. a) Let f x  x5 and gx  x2  3x 1 find  f  g'x Solution: Because f x  x5 then f 'x  5x4 , thus f 'gx  5gx4  f 'gx  5 x2  3x 1 4 , furthermore Because gx  x2  3x 1 then g'x  2x  3 Therefore  f  g'x  f 'gxg'x  f  g'x  5 x2  3x 1 4 2x  3 Illustration 2.9. b) Let f x  x3 and gx  2 find f  g'x x 1 Solution: Because f x  x3 then f 'x  3x2 , thus f 'gx  3gx2 f 'gx  3 2 2 , furthermore  x 1 Because gx  2 then g'x  2 x 1 x 12 Therefore  f  g'x  f 'gxg'x 57

MATH 401 – DIFFERENTIAL CALCULUS g 'x  3 2  2  2    f   x 1 x 12 f  g'x   12  2  x 12 x 12 f  g 'x   24 x 14 Theorem 2.9. The Chain Rule of Differentiation If g is differentiable at x and if f is differentiable at g(x), then the composition f  g is differentiable at x. Moreover, if y=f(g(x)) and u=g(x) then y=f(u) and dy  dy  du dx du dx In words, the derivative of a composite function of f and g is equal to the product of their respective derivatives. Illustration 2.10. a) Given y  u10 and, u  2x3  5x2  4 find dy dx Solution: from y  u10 , the derivative is dy  10u9 du from u  2x3  5x2  4 , the derivative is du  6x2 10x , then dx  dy dy  du 6x2 du dx dx   10u9 10x , since u  2x3  5x2  4 , then    dy  10 2x3  5x2  4 9 6x2 10x dx b) Given y  5u3  2 and, u  3  x4 find dy dx Solution: from y  5u3  2 , the derivative is dy  15u2 du from u  3  x4 , the derivative is du  4x3 , then dx 58

MATH 401 – DIFFERENTIAL CALCULUS  dy  dy  du  15u2  4x3  60u2 x3 , since u  3  x4 , then dx du dx  dy  60x3 3  x4 2 dx Theorem 2.10. If x  f t and y  gt are parametric equations, the derivative of y with respect to x is equal to the quotient of the derivative of y with respect to t divided by the derivative of x with respect to t . dy  dy dt dx  0 dx dx dt ; dt Illustration 2.11. dy a) x  2t 2  4t and y  4t  2 , find dx Solution: from x  2t 2  4t , the derivative is dx  4t  4 dt from y  4t  2 , the derivative is dy  4 , then dt dy  dy dt  4 4  4 dx dx dt 4t  4t 1 dy  t 1 dx 1 b) If x3 and y  2t 1 , find dy t 4 dx from x 3 , the derivative is dx  3 t dt t2 from y  2t 1 , the derivative is dy  1 , then 4 dt 2 1 dy  dy dt  2   1  t2  dx dx dt 3  2 3 t2 59

MATH 401 – DIFFERENTIAL CALCULUS dy  t 2 dx 6 Theorem 2.11. If the variable x is a function of y then the derivative dy dx is equal to the reciprocal of the derivative of the given function with respect to y . If x  gy , then dy  1 dx dx dy Illustration 2.12. a) Given x  3y3 , find dy dx Solution: from x  3y3 , the derivative is dx  9 y2 dy dy  1  1 , from x  3y3 the value of y  3 9x dx dx 9y2 3 dy  dy  1  1  1  1 9y2 81x2 dx 9x 3 9x 2  3 9 2 3 9 9  3      dy  1 3 9x  3 9x  dx 3 81x2 3 9x 3 729x3 dy  3 9x dx 9x Theorem 2.12. If u is a differentiable function of x, then du d  u  dx dx 2 u The derivative of a radical whose index is two, is a fraction whose numerator is the derivative of the radicand, and whose denominator is 60

MATH 401 – DIFFERENTIAL CALCULUS twice the given radical, if the derivative exists Illustration 2.13. a) Given f x  2x2  4x , find f 'x Solution: d 2x2  4x 2x2  4x  dx  f 'x  d dx 2 2x2  4x  4x  4 2 2x2  4x  22x  2 2 2x2  4x f 'x  2x  2 2x2  4x b) If y  3  2x2 find y' d 3  2x2 3  2x2  dx  y' d dx 2 3  2x2  0  4x 2 3 2x2   4x 2 3  2x2 y'  2x 3  2x2 61

MATH 401 – DIFFERENTIAL CALCULUS Exercise2. Differentiate the following function by applying the theorems 1. f x  1 x2  4x  9 Answer: x+4 Answer: 40x3 15x2 - 20 2  Answer: x4  5x2  2 2. f x  2  x3 10x  5 x2 1 2 3. y  x3  2x x2 1 4. y   5 3 Answer:  375  x 1 x 14 5. If f x  4x2 and gx  3x2  1 find  f  g'x  Answer: 48x 3x2 1 6. If y  u2  9 and u  3x2  5 find dy  Answer: 12x 3x2  5 dx 7. If x t and y t2 , find dy Answer: t dx t  4 t  42 4t  4 8. If x  2y2  y  4 , find dy Answer: 1 dx  184  x 9. If f x  x 3  2x2 , find f 'x Answer:  4x2  3 3 2x2 62

MATH 401 – DIFFERENTIAL CALCULUS 2.4 Higher-Order Derivatives The derivative f ' of a function f is itself a function and hence may have a derivative of its own. If f ' is differentiable, then its derivative is denoted by f '' and is called the second derivative of f . As long as we have differentiability, we can continue the process of differentiating to obtain the third, fourth, fifth, and even higher derivatives of f . These successive derivatives are denoted by f ', f ''  ( f ')', f '''  ( f '')', f 4  ( f ''')', f 5  ( f 4 )',... Using the same functional notation y  f (x) , we will have the following symbols for higher derivatives: First Derivative: y' , f '(x) , dy d  y , d  f  x , Dx f x dx , dx dx Second Derivative: y'' , f ''(x) , d2y d2  y , d2  f  x , D2x f x dx2 dx2 , dx2 Third Derivative: y''' , f '''(x) , d3y d3  y , d3  f  x , D3x f  x dx3 , dx3 dx3 nth Derivative: yn , f n (x) , dny dn  y , dn  f  x , Dnx f x dxn , dxn dxn The symbols dy , d2y d n y are called Leibniz notations. dx dx2 , dxn Example1. Given f x  x3  2x2  3x  5 , find f '(x) , f ''(x) and f '''(x) Solution: f 'x  3x2  4x  3 f ''x  6x  4 f '''x  6 Example2. Find f '(x) , f ''(x) and f '''(x) given that f x  x2  2x1 Solution:  f 'x  2x  2  x2  2x  2x2 63

MATH 401 – DIFFERENTIAL CALCULUS  f ''x  2  2  2x3  2  4x3  f '''x  0  4  3x4  12x4 Example3. Find the first and second derivatives of f x  x2 x  5x Solution: f x  x2 x  5x  f x  x2 x1 2  5x  f x  x5 2  5x f 'x  5 5 1  5  5 x3 2  5 x2 22 f ' ' x  5  3 3 1   0 2 2 x2 f ' ' x  15 x1 2or 15 x 44 Exercise3. Determine the first and second derivative given the following functions. 1. f x  x3  9x2  27x  27 Answer: 3x2 18x  27 , 6x 18 2. y  1 Answer: 1 , 2 x x2 x3 3. f x  4 x2 1 2 Answer: 16x3 16x , 48x2 16  4. y  1 x2  3x 2 Answer: 3x2 12x  9 , 6x 12 x 64

MATH 401 – DIFFERENTIAL CALCULUS 5. f x  9  x2  Answer:  x ,  9 9 x2 9  x2 3 2 2.5 Implicit Differentiation There are two ways to define functions, implicitly and explicitly. Most of the equations we have dealt with have been explicit equations, such as y  2x  3 , so that we can write y  f (x) where f (x)  2x 3 . But the equation 2x  y  3 describes the same function. This second equation is an implicit definition of y as a function of x . As there is no real distinction between the appearance of x or y in the second form, this equation is also an implicit definition of x as a function of y . An implicit function is a function in which the dependent variable has not been given \"explicitly\" in terms of the independent variable. To give a function f explicitly is to provide a prescription for determining the output value of the function y in terms of the input value x: y  f (x) . By contrast, the function is implicit if the value of y is obtained from x by solving an equation of the form: f (x, y)  0 . An equation of the form y  f (x) is said to define explicitly as a function of x because the variable y appears alone on one side of the equation and does not appear at all on the other side. However, sometimes functions are defined by equations in which y is not alone on one side; for example the equation yx  y 1 x is not of the form y  f (x) , but still defines y as a function of x since it can be rewritten as y  x 1 . Thus we say that yx  y 1 x defines y x 1 implicitly as a function of x , the function being f (x)  x 1 . x 1 Suppose we have an equation f (x, y)  0 where neither variable could be expressed as a function of the other. In other words, it wouldn’t be possible, by rearranging f (x, y)  0 , to separate out one of the variables and express it as a function of the other. Often we can solve an equation f (x, y)  0 for one of the variables obtaining multiple Solutions constituting multiple branches. Consider the equation x2  y2 1 0 which defines y as an implicit function of x . If we solve for y in terms of x , we obtain two Solutions y   1 x2 and y   1 x2 thus we have found two functions that are defined implicitly by x2  y2 1 0 . 65

MATH 401 – DIFFERENTIAL CALCULUS In general, it is not necessary to solve an equation for y in terms of x in order to differentiate the functions defined implicitly by the equation. To find the derivative of functions defined implicitly we use implicit differentiation. Steps in Implicit Differentiation: 1. Differentiate both sides of the equation with respect to x . 2. Collect all the terms with dy on one side of the equation. dx 3. Factor out dy and solve for it. dx Example1. Given x2  4y  5, find dy . dx Step1. 2x  4 dy  0 dx Step2. 4 dy  2x dx Step3. dy   2x  dy   1 x dx 4 dx 2 Example2. Given x3  y3  5xy  0, solve for dy dx    Step1. d x3  d y3  d 5xy  d 0 dx dx dx dx 3x2  3y2 dy  5 x dy  y1  0 dx  dx  3x2  3y2 dy  5x dy  5y  0 dx dx Step2. 3y2 dy  5x dy  3x2  5y dx dx      Step3. dy 3y2  5x   3x2 5y  dy   3x2  5y dx dx 3y2  5x Example3. Given 2x2  x2 y2  y3  4 , find y' Step1. 4x  [(x2 )(2yy' )  ( y2 )(2x)]  (3y4 y' )  0 4x  2x2 yy'2xy2  3y4 y'  0 66

MATH 401 – DIFFERENTIAL CALCULUS Step2.  2x2 yy'3y4 y'  4x  2xy2      Step3.  2x2 y  3y4 y'  4x  2xy2  y'   4x  2xy 2  2x2 y  3y4 Exercise4. Find dy or y' by implicit differentiation dx 1. 4x2  9y2  36  4x Answer: 9 y 2. x3  2x2  y4  y  5 3x2  4x  Answer: 4 y3 1 3. 4x3  5x2 y2  4y3  3y  12x2 10xy2 4. x3 y3  3xy  Answer: 10x2 y 12y2  3  y  x2 y3  Answer: x3 y2  x 5. xy2  2y3  3xy  2x2  y3  3y2 4x  y2  3y  Answer: 2xy  3y2  3x  6y 67

MATH 401 – DIFFERENTIAL CALCULUS CHAPTER TEST (Problem Set 2). Solve the following completely and neatly as possible. Find the derivative of the following using the three-step rule 1. y  3x 1 2x 5 2. y  a2  x2 Differentiate the following function by applying the theorems on differentiation of algebraic functions   3. ht  t4  3t2  5 3t4  t2 4. y  3x  2 2 2x  6 2 5. x  t 2 2 and y  t t4 t  42 6. y  u2  9 and u  4x3  9 Find the first, second and third derivative of the given function 7. f x   t  4  22 8. y  1  x x Find y’ using implicit differentiation given the following functions 9. x  y  xy  21 10. x2 y2  x2  y2 68

MATH 401 – DIFFERENTIAL CALCULUS CHAPTER III DERIVATIVES OF TRANSCENDENTAL FUNCTIONS So far you have studied one of the elementary functions – the algebraic functions. In this chapter, the learners will study the derivatives of the remaining elementary functions – the transcendental functions. Theorems presented in this chapter are taken from [1] Larson, R. (2018), [2] Leithold, L. (2002) and [3] Stewart, J. (2016) At the end of this chapter, the student should be able to: 1. Define the derivatives of transcendental functions such as trigonometric, inverse trigonometric, logarithmic, exponential and hyperbolic functions. 2. Apply the theorems of differentiation in finding the derivatives of transcendental functions. 3.1 TRIGONOMETRIC FUNCTIONS With this section we are going to start looking at the derivatives of functions other than polynomials or roots of polynomials. We will start this process off by taking a look at the derivatives of the six trigonometric functions. The basic six trigonometric functions include the following six functions: sine (sin x), cosine (cos x), tangent (tan x), cotangent (cot x ), secant (sec x), and cosecant (csc x). All these functions are continuous and differentiable in their domains. Before we actually get into the derivatives of the trigonometric functions we need to recall a couple of limits that will show up in the derivation of two of the derivatives. lim lim 3.1.1. Derivatives of Trigonometric Functions The derivatives of the six trigonometric functions are summarized in the following theorem. Theorem 3.1. Derivatives of Trigonometric Functions 69

MATH 401 – DIFFERENTIAL CALCULUS Proof: Let f be the sine function, so that From the definition of the derivative of a function, the derivative for the sine function can be written as To evaluate this limit, we will need to use the trigonometric identity For the proof of the derivative of cosine function, we can follow a similar process Let g be the cosine function, so that 70

MATH 401 – DIFFERENTIAL CALCULUS From the definition of the derivative of a function, the derivative for the cosine function can be written as To evaluate this limit, we will need to use the trigonometric identity The derivatives of tangent, cotangent, secant and cosecant functions are obtained from the trigonometric identities involving sine and cosine functions as well as the above derivatives of the sine and cosine functions, and the previous theorems on differentiation of algebraic functions. For the tangent function, we will use the trigonometric identity Applying the Quotient Rule of differentiation, we have 71

MATH 401 – DIFFERENTIAL CALCULUS For the cotangent function, we will use the trigonometric identity Applying the Quotient Rule of differentiation, we have For the secant function, we will use the trigonometric identity 72

MATH 401 – DIFFERENTIAL CALCULUS Applying the Quotient Rule of differentiation, we have For the cosecant function, we will use the trigonometric identity Applying the Quotient Rule of differentiation, we have 73

MATH 401 – DIFFERENTIAL CALCULUS Let us have the following examples: Example 3.1.1.1 Differentiate Solution: We will need to use the Product Rule of differentiation on the first term. Also the constant 2 will be considered part of the first function in the product of the first term. Here is the derivative of the function y Example 3.1.1.2 Find the derivative of Solution: We will need to use the Product Rule of differentiation on the second term. Be careful with the minus sign in front of the second term and make sure it gets dealt with properly. There are two ways to deal with this. One way is to make sure that you use a set of parentheses as follows: Another way to do this is to consider minus sigh as part of the first function in the product of the second term. Doing this gives 74

MATH 401 – DIFFERENTIAL CALCULUS So regardless how you approach this problem, you will get the same derivative. Example 3.1.1.3 Find the derivative of at . Solution: Just differentiate each term using the formula above. At . the value of the first derivative of h(x) is Example 3.1.1.4 Differentiate Solution: We will use the quotient rule of differentiation to evaluate the derivative of this function. We can still simplify this by factoring out 3 in the last two terms in the numerator and use the Pythagorean identity 75

MATH 401 – DIFFERENTIAL CALCULUS Since the numerator is exactly a factor of the denominator, we can canceled them and gives Example 3.1.1.5 Differentiate Solution: We need to use the product rule of differentiation for both terms. Combining like terms, we have 3.1.2. Chain Rule of the Derivatives of Trigonometric Functions In this section, we will determine the derivatives of trigonometric functions using the Chain Rule. This rule deals with composite functions involving trigonometric functions. The Chain Rule versions of the derivatives of the six trigonometric functions are as follows: Let u be a differentiable function of x 76

MATH 401 – DIFFERENTIAL CALCULUS Example 3.1.2.1 Apply the Chain Rule to find the derivative of the following functions: (a) (b) Solution: Solution: (c) (d) Solution: Solution: (e) (f) Solution: Solution: Example 3.1.2.2 To understand the mathematical conventions regarding parentheses, apply the Chain Rule to find the derivative of the following functions: Function Derivative (a) (b) (c) (d) 77

MATH 401 – DIFFERENTIAL CALCULUS (e) Example 3.1.2.3 Find the derivative of the following: (a) Solution: We must use the Product Rule and Chain Rule. (b) We must use the Quotient Rule along with the Chain Rule. Solution: (c) Solution: Recognize here that we have function that is inside the function; that is . We will start using the General Power Rule, then the Chain Rule by approaching this step-by-step. 78

MATH 401 – DIFFERENTIAL CALCULUS To express this in terms of trigonometric function only, we can use the Pythagorean identity 3.1.3. Higher-Order Derivatives of Trigonometric Functions The higher-order derivatives are denoted as follows: First Derivative : Second Derivative: Third Derivative: Fourth Derivative: nth Derivative: Let’s have the following examples: Example 3.1.3.1 Find the second derivative of the following: (a) (b) Solution: Solution: () (d) Solution: Solution: 79

MATH 401 – DIFFERENTIAL CALCULUS (d) Solution: Using the Pythagorean identity Example 3.1.3.2 Find the third derivative of the following: at Solution: at (b) Solution: Use the Power Rule and Chain Rule Use the Product Rule, Power Rule and Chain Rule Using the Pythagorean identity Combining like terms, we have 80

MATH 401 – DIFFERENTIAL CALCULUS As an alternative, the final answer can be expressed in terms of tangent function only. (c) Solution: Use the Product Rule, Power Rule, and Chain Rule Use the Product Rule, Power Rule, and Chain Rule in each term (d) Solution: Using the Pythagorean identity Its third derivative is 81

MATH 401 – DIFFERENTIAL CALCULUS Expressing this in terms of one function only, use the Pythagorean identity As an alternative, the final answer can be expressed in terms of cotangent functions only. Example 3.1.3.3 Find the 4th derivative of the following: (a) (b) Solution: Solution: Remarks: The higher-order derivatives of sin x and cos x follow a repeating pattern. Once we recognize the pattern of derivatives, we can find any higher-order derivative by determining the step in the pattern to which it corresponds. For example, every fourth derivative of sin x equals sin x , so while The pattern of derivatives is that the first derivative of sin x equals its 5th derivative. The 2nd derivative of sin x equals its 6th derivative. The nth derivative of the function . For n greater than 4, we can perform division algorithm n/4 = q + r/4, where q is the quotient and r is the remainder. It is now evident that for the nth derivative of sin x, , it follows that , where . Example 3.1.3.4 Evaluate the indicated nth derivative of sin x. 82

MATH 401 – DIFFERENTIAL CALCULUS (a) Find the 50th derivative of sin x. Solution: Using the algorithm, 50 = 4(12) + 2 Since r = 2, then (b)Find the 111th derivative of sin x. Solution: Using the algorithm, 111 = 4(27) + 3 Since r = 3, then (c)Find the 1200th derivative of sin x. Solution: Using the algorithm, 1200 = 4(300) + 0 Since r = 0, then The same pattern for the nth derivative of the function which is . Example 3.1.3.5 Evaluate the indicated nth derivative of cos x. (a)Find the 25th derivative of cos x. Solution: Using the algorithm, 25 = 4(6) + 1 Since r = 1, then (b)Find the 335th derivative of cos x. Solution: Using the algorithm, 335 = 4(83) + 3 Since r = 3, then Example 3.1.3.6 Find the 4th derivative of the following: 83

MATH 401 – DIFFERENTIAL CALCULUS (a) (b) Solution: Solution: x Remarks: A similar pattern of higher-order derivatives can be observed for the functions and , where is a constant. The only difference is the numerical coefficient which is equal to . For instance, the 4th derivative of is . In (b), the 3rd derivative of is . Example 3.1.3.7 Evaluate the indicated nth derivative: (a) Find the 99th derivative of Solution: Using the algorithm, 99 = 4(24) + 3 Since r = 3 and then (b)Find the 77th derivative of Solution: Using the algorithm, 77 = 4(19) + 1 Since r = 1 and then 84

MATH 401 – DIFFERENTIAL CALCULUS 3.1.4. Implicit Differentiation of Trigonometric Functions In this section, we will apply the process of implicit differentiation for equations involving trigonometric functions. Implicit differentiation is a technique based on a Chain Rule that is used to find the derivative when the relationship between the variables is given implicitly rather than explicitly (solved for one variable in terms of the other). Example 3.1.4.1 Find the given that (a) . Solution: Since it is difficult to express y in terms of x only, we will differentiate implicitly with respect to x. (b) Solution: Since it is difficult to express y in terms of x only, we will differentiate implicitly with respect to x. 85

MATH 401 – DIFFERENTIAL CALCULUS Example 3.1.4.2 Find the given that (a) Solution : Using the result of of the Example 3.1.4.1 (b), we can evaluate the 2nd derivative as follows By implicit differentiation with respect to x, we have Substitute the expression of (b) Solution: Start with the implicit differentiation with respect to x To evaluate the , perform another implicit differentiation with respect to x. Use Product Rule on the right side. 86

MATH 401 – DIFFERENTIAL CALCULUS Substitute the expression for y’ Exercise 3.1. Find the derivative of the following functions: 1. ans. 2. at ans. 3. ans. 4. at ans. 5. ans. 6. ans. Find the indicated nth derivative of the following: 7. 250th derivative of ans. 8. 85th derivative of ans. Use implicit differentiation to find the ans. 9. of ans. 10. of sin 87

MATH 401 – DIFFERENTIAL CALCULUS 3.2. Inverse Trigonometric Functions In this section, we are to explore the process of finding the derivatives of the six inverse trigonometric functions. 3.2.1. Derivatives of Inverse Trigonometric Functions The following theorem lists the derivatives of the six inverse trigonometric functions. Theorem3.2 Derivatives of Inverse Trigonometric Functions Let u be a differentiable function of x. It can be noted that the derivative of Arccos u, Arccot u, and Arccsc u are the negatives of the derivatives of Arcsin u, Arctan u, and Arcsec u, respectively. Proof for Arcsin u: Let . So, , and you can use implicit differentiation with respect to x as follows. Then use the Pythagorean identity So 88

MATH 401 – DIFFERENTIAL CALCULUS If u is a differentiable function of x, then you can use the Chain Rule to write Proof for Arctan u: Let . So, , and you can use implicit differentiation with respect to x as follows. Then use the Pythagorean identity If u is a differentiable function of x, then you can use the Chain Rule to write Proof for Arcsec u: . So, , and you can use Let implicit differentiation with respect to x as follows. Then use the Pythagorean identity 89

MATH 401 – DIFFERENTIAL CALCULUS If u is a differentiable function of x, then you can use the Chain Rule to write Let us analyze the following examples: Example 3.2.1.1 Find the derivative of the following functions: (a) (d) Solution: , then Solution: , then Let Let (b) , then (e) Solution: Solution: Let Let , then (c) , then (f) Solution: Solution: since the derivative of Arccsc u if Let 90

MATH 401 – DIFFERENTIAL CALCULUS the negative of that of Arcsecu, then Example 3.2.1.2 Find the derivative of Solution: Apply the Product Rule on the second term. 3.2.2 Chain Rule of the Derivatives of Inverse Trigonometric Functions In this section, we will determine the derivatives of inverse trigonometric functions using the Chain Rule. This rule deals with composite functions involving inverse trigonometric functions. Example 3.2.2.1 Find the derivative of the following function: (a) Solution: (b) 91

MATH 401 – DIFFERENTIAL CALCULUS Example 3.2.2.2 Use Chain Rule to find the derivative of the following: (a) Solution: (b) Solution: 3.2.3 Higher-Order Derivatives of Inverse Trigonometric Functions In this section, we are to evaluate the higher-order derivatives of inverse trigonometric functions. Example 3.2.3.1 Find the second derivative of Solution: We have determined in Example 3.2.2.1 (a) its first derivative as Now to find its 2nd derivative, apply the Quotient Rule 92

MATH 401 – DIFFERENTIAL CALCULUS Example 3.2.3.2 Find the third derivative of Solution: Its first derivative with respect to x is Then its second derivative with respect to x is So the third derivative is 3.2.4 Implicit Differentiation of Inverse Trigonometric Functions In this section, we will apply the process of implicit differentiation for equations involving inverse trigonometric functions. Example 3.2.4.1 Use implicit differentiation to find the derivative of Solution: 93

MATH 401 – DIFFERENTIAL CALCULUS Example 3.2.4.2 Use implicit differentiation to find the derivative of Solution: 94

MATH 401 – DIFFERENTIAL CALCULUS Exercise 3.2. 1. Verify each differentiation formula (a) (b) (c) 2. True or False. The for all x in the domain. ans. True Find the derivative of the following functions: ans. 3. ans. 4. ans. 5. ans. 6. Find the indicated nth derivative of the following: 7. 2nd derivative of ans. 8. 5th derivative of ans. Use implicit differentiation to find the derivative of: 9. ans. 10. y ans. 95

MATH 401 – DIFFERENTIAL CALCULUS 3.3. Logarithmic Functions The definition of the logarithmic function that you encountered in Chapter 1 was based on exponents, and the properties of logarithms were then proved from the corresponding properties of exponents. In this section, we will consider the derivatives of the natural logarithmic function, denoted by ln, as well as the derivative of logarithmic function to base . 3.3.1. Derivatives of Logarithmic Functions The derivative of the logarithmic function is determined by the following theorem. Theorem 3.3 Chain Rule of the Derivative of Logarithmic Function Let be a positive real number and let u be a differentiable function of x and , then The second formula is a special case of the first formula since the natural logarithm . In the first formula we can set thus . Example 3.3.1.1 Find the derivative of the following functions: (a) y where is constant Solution: Solution: y y y y (b) y Solution: y y 96

MATH 401 – DIFFERENTIAL CALCULUS (c) Another Solution: Rewrite the given using a logarithmic property Solution: y , y y y (d) y where Solution: Solution: y y y y y (e) y y Solution: Take note that Rewrite the given as: y Use the Power Rule and Chain Rule to evaluate its derivative y Example 3.3.1.2 Find the derivative of the following functions: (a) y 0 Solution: where are constants Solution: y ln 0 y (b) y Solution: y y ln 97