diff cal

MATH 401 – DIFFERENTIAL CALCULUS y ln y 3.3.2. Higher-Order Derivatives of Logarithmic Functions In this section, we will evaluate the nth derivatives of logarithmic functions. Example 3.3.2.1 Find the second derivative of the following functions: (a) y Solution: y y 1x x (b) Solution: y (c) Solution: y Example 3.3.2.2 Find the third derivative of the following functions: (a) Solution: 98

MATH 401 – DIFFERENTIAL CALCULUS (b) . Solution: We can rewrite the given as (c) Solution: 3.3.3. Implicit Differentiation of Logarithmic Functions In this section, we will apply the process of implicit differentiation for equations involving logarithmic functions. Example 3.3.3.1 Use implicit differentiation to find the derivative of (a) . Solution: We can rewrite the second term applying the logarithmic property and 99

MATH 401 – DIFFERENTIAL CALCULUS 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. 100

MATH 401 – DIFFERENTIAL CALCULUS 3.3.4. Logarithmic Differentiation In this section we will evaluate the derivatives of some complicated functions by using logarithms. That process is called logarithmic differentiation which was developed in 1697 by Johann Bernoulli(1667- 1748). Let’s see how this works in the following examples: Example 3.3.4.1 Find the derivative of the function Solution: The derivative of this function can be evaluated by using the Product Rule and Quotient Rule but it is somewhat a complicated process. To simplify the process, we can take the logarithms of both sides. We need to use the properties of logarithms to expand the right side as follows. Perform implicit differentiation, we have Multiply both sides by y and substitute the expression for y, we have 101

MATH 401 – DIFFERENTIAL CALCULUS This can still be simplified using algebraic rules. Example 3.3.4.2 Find the derivative of the function Solution: The derivative of this function can be evaluated by using the Product Rule Power Rule and Chain Rule but it is somewhat a complicated process. To simplify the process, we can take the logarithms of both sides. Use the properties of logarithms to expand the right side as follows. Perform implicit differentiation, we have Multiply both sides by f(z) and substitute the expression for f(z), we have Simplifying this algebraically 102

MATH 401 – DIFFERENTIAL CALCULUS Exercise 3.3. Find the derivative of the following function: 1. ans. 2. ans. 3. ans. 4. ans. 5. ans. 6. ans. Find the indicated nth derivative of the following: 7. 2nd derivative of ans. 8. 5th derivative of ans. Use implicit and/or logarithmic differentiation to find the derivative y’ of: 9. ans. 10. ans. 103

MATH 401 – DIFFERENTIAL CALCULUS 3.4. Exponential Functions Because the natural logarithmic function is increasing on its entire domain, then by the inverse function theorem, it has an inverse that is also an increasing function. The inverse of ln is called the natural exponential function, denoted by exp. It is defined by if and only if . In general, the exponential function to the base is defined by , where is any positive number and x is any real number. In this section, we will explore the derivatives of these exponential functions. 3.4.1. Derivatives of Exponential Functions Theorem 3.4.1 Derivatives of Exponential Functions Let be a positive real number Proof : then applying logarithms on both sides, we have Let Apply implicit differentiation with respect to x on both sides Multiply both sides by y Replacing y by , we obtain The proof of the first formula is given as follows: From the relationship between exponential and logarithmic function that , we set and using the property of logarithm , then Differentiating both sides with respect to x, we obtain 104

MATH 401 – DIFFERENTIAL CALCULUS Example 3.4.1.1 Derivatives Exponential Functions Before moving on to the next section we need to be aware of the distinction between these two derivatives: It is important to note that with the Power Rule the exponent n MUST be constant and the base x MUST be a variable while for the derivative of an exponential function, the exponent x MUST be a variable and the base must be a constant. In cases where both the exponent and the base involve variables will be considered in a later section. 3.4.2. Chain Rule of the Derivatives of Exponential Functions Theorem 3.4.2 Chain Rule of Derivatives of Exponential Functions Let be a positive real number and let u be a differentiable function of x Observe that the derivative of the function defined by , where k is a constant, is itself. The only other function we have previously encountered that has this property is the constant function zero; actually, this is the special case of when . 105

MATH 401 – DIFFERENTIAL CALCULUS Example 3.4.2.1 Apply the Chain Rule to find the derivative of the following functions: (b) Solution: Solution: Solution: (d) Solution: (e) (f) Solution: Solution: Use the property of exponent Differentiate both sides Example 3.4.2.2 Find the derivative of the following functions: y where are constants Solution: Solution: k y ln y y ln y Or y (b) y 0 y Solution: y ln 10 0 yk y ln 10 0 y ln 10 0 106

MATH 401 – DIFFERENTIAL CALCULUS 3.4.3. Higher-Order Derivatives of Exponential Functions In this section, we will evaluate the nth derivatives of exponential functions. Example 3.4.3.1 Find the second derivative of the following functions: Solution: (b) Solution: Example 3.4.3.2 Find the nth derivative of , m is constant. Solution: So the nth derivative of can be written as 3.4.4. Implicit Differentiation of Exponential Functions In this section, we will apply the process of implicit differentiation for equations involving exponential functions. Example .4.4.1 Use implicit differentiation to find the derivative y’ of (a) . Solution: We can rewrite the 3rd term applying the rules of exponents Since it is difficult to express y in terms of x only, differentiate implicitly with respect to x. 107

MATH 401 – DIFFERENTIAL CALCULUS Combining like terms (b) Solution: Since it is difficult to express y in terms of x only, differentiate implicitly with respect to x. Combining like terms 3.4.5. Logarithmic Differentiation Logarithmic differentiation can also be used to evaluate the derivative of function of this form: These are the cases where both the exponent and the base involve variables. Example 3.4.5.1 Differentiate the function We have encountered the derivative of two similar functions like this, 108

MATH 401 – DIFFERENTIAL CALCULUS But neither of the two will work here because both the base and the exponent are variables. Logarithmic differentiation can be used in this case. To simplify the process, we can take the logarithms of both sides. Use the properties of logarithms to expand the right side as follows. Perform implicit differentiation, we have Multiply both sides by y and substitute , we have Now let us consider a more complicated example of this. Example 3.4.5.2 Differentiate the function Solution: Again the Power Rule and the derivative of exponential functions will not work here because both the base and the exponent are variables. Logarithmic differentiation can be used in this case. To simplify the process, we can take the logarithms of both sides. Use the properties of logarithms to expand the right side as follows. Perform implicit differentiation, we have 109

MATH 401 – DIFFERENTIAL CALCULUS Multiply both sides by y and substitute , we have Example 3.4.5.3 Differentiate the function Solution: Again the Power Rule and the derivative of exponential functions will not work here because both the base and the exponent are variables. Logarithmic differentiation can be used in this case. To simplify the process, we can take the logarithms of both sides. Use the properties of logarithms to expand the right side as follows. Perform implicit differentiation, we have Multiply both sides by y and substitute , we have 110

MATH 401 – DIFFERENTIAL CALCULUS Exercise 3.4. Find the derivative of the following exponential functions: 1. ans. 2. ans. 3. ans. Find the indicated nth derivative of the following functions: 4. 2nd derivative of ans. 5. 100th derivative of ans. Use implicit differentiation to find the derivative y’ of the following: 6. 0 ans. 7. 0 0 ans. Use logarithmic differentiation to find the derivative of the following: 8. ans. 9. ans. 10. ans. 111

MATH 401 – DIFFERENTIAL CALCULUS 3.5. Hyperbolic Functions Because hyperbolic functions can be written in terms of the exponential functions, you can easily derive rules for their derivatives. In this section, we will explore the derivatives of the hyperbolic functions. 3.5.1. Derivatives of Hyperbolic Functions Theorem3.5. Derivatives of Hyperbolic Functions d sinhx  coshx d coshx  sinhx dx dx d tanhx  sec h2 x d cothx   csc h2 x dx dx d sec hx  sec hxtanhx d csc hx  csc hxcothx dx dx Proof: Recall the definition of hyperbolic sine function Differentiate both sides with respect to x Next, for the hyperbolic tangent function, it is defined as Differentiate both sides with respect to x 112

MATH 401 – DIFFERENTIAL CALCULUS Use Quotient Rule on the right side Use the hyperbolic identity Let’s have the following examples: Example 3.5.1.1 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 3.5.1.2 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: Example 3.5.1.3 Find the derivative of . Solution: Just differentiate each term using the formula above. 113

MATH 401 – DIFFERENTIAL CALCULUS Example 3.5.1.4 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 hyperbolic identity 114

MATH 401 – DIFFERENTIAL CALCULUS 3.5.2. Chain Rule of the Derivatives of Hyperbolic Functions In this section, we will determine the derivatives of hyperbolic functions using the Chain Rule. This rule is applied on differentiating composite functions involving hyperbolic functions. The Chain Rule versions of the derivatives of the six hyperbolic functions are as follows: Let u be a differentiable function of x Example 3.5.2.1 Apply the Chain Rule to find the derivative of the following functions: (b) Solution: Solution: Solution: (d) Solution: (e) (f) Solution: Solution: 115

MATH 401 – DIFFERENTIAL CALCULUS Example 3.5.2.2 To understand the mathematical conventions regarding parentheses, apply the Chain Rule to find the derivative of the following functions: Solution: Derivative Function (a) Example 3.5.2.3 Find the derivative of the following: Solution: We must use the Product Rule and Chain Rule. Solution: We must use the Quotient Rule along with the Chain Rule. 116

MATH 401 – DIFFERENTIAL CALCULUS 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. To express this in terms of hyperbolic function only, we can use the hyperbolic identity Solution: Solution: Solution: 117

MATH 401 – DIFFERENTIAL CALCULUS 3.5.3. Higher-Order Derivatives of Hyperbolic Functions Let’s have the following examples: Example 3.5.3.1 Find the second derivative of the following: (b) Solution: Solution: () (d) Solution: Solution: (e) Solution: Expressing this in terms of hyperbolic secant only, we can use the hyperbolic identity (d) . Use Chain Rule Solution: This a composite function, where 118

MATH 401 – DIFFERENTIAL CALCULUS Then differentiate to get the second derivative Example 3.5.3.2 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 13 13 13 Using the hyperbolic identity Combining like terms, we have 119

MATH 401 – DIFFERENTIAL CALCULUS (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: Use the Power Rule and Chain Rule Using the hyperbolic identity Differentiate each term to get the third derivative Expressing this in terms of one function only, use the hyperbolic identity 120

MATH 401 – DIFFERENTIAL CALCULUS Combining like terms, we have As an alternative, the final answer can be expressed in terms of hyperbolic cotangent functions only. Example 3.5.3.3 Find the 4th derivative of the following: Solution: (b) Solution: Remarks: The higher-order derivatives of sinh x and cosh 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 second derivative of sinh x equals sinh x , so while The pattern of nth derivative for hyperbolic sine function sinh x is For hyperbolic cosine function cosh x, the pattern for nth derivative is Let us have the following examples: 121

MATH 401 – DIFFERENTIAL CALCULUS Example 3.5.3.4 Evaluate the indicated nth derivative of sinh x. (a) Find the 50th derivative of sinh x. Solution: Using the pattern, since n = 50 and it is even, then (b)Find the 111th derivative of sinh x. Solution: Using the pattern, since n = 111 and it is odd, then Example 3.5.3.5 Evaluate the indicated nth derivative of cosh x. (a)Find the 25th derivative of cosh x. Solution: Using the pattern, since n = 25 and it is odd, then (b)Find the 3,350th derivative of cosh x. Solution: Using the pattern, since n =3.350 and it is even, then 3.5.4. Implicit Differentiation of Hyperbolic Functions In this section, we will apply the process of implicit differentiation for equations involving hyperbolic 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 3.5.4.1 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. 122

MATH 401 – DIFFERENTIAL CALCULUS Combining like terms, (b) Solution: Since it is difficult to express y in terms of x only, we will differentiate implicitly with respect to x. Example 3.5.4.2 Find the given that (a) Solution : Using the result of of the Example 3.5.4.1 (b), we can evaluate the 2nd derivative as follows By implicit differentiation with respect to x, we have 123

MATH 401 – DIFFERENTIAL CALCULUS 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 and Chain Rule on the right side. Substitute the expression for y’ 124

MATH 401 – DIFFERENTIAL CALCULUS Exercise 3.5. Verify the following differentiation formula: 1. 2. 3. 4. Find the derivative of the following functions: 5. ans. 6. ans. 7. ans. Find the indicated nth derivative of the following: 8. 25th derivative of ans. 9. 44th derivative of ans. Use implicit differentiation to find the derivative of 10. ans. 125

MATH 401 – DIFFERENTIAL CALCULUS Problem Set No. 3 Derivatives of Transcendental Functions Find the derivative of the following functions: 1. 2. 3. 4. 5. 6. Find the indicated nth derivative of the following: 7. 222nd derivative of 8. 115th derivative of Use implicit and/or logarithmic differentiation to find the derivative y’ of: 9. 10. 126

MATH 401 – DIFFERENTIAL CALCULUS CHAPTER IV APPLICATIONS OF DERIVATIVES OF ALGEBRAIC AND TRANSCENDENTAL FUNCTIONS We have learned from Chapter II and Chapter III how to find the derivatives of algebraic and transcendental functions by applying both the definitions and theorems of differentiation. This chapter discusses several applications of derivatives. We shall now learn the concepts and the process of solving problems involving techniques on approximation, tangent line and normal line to a given curve, curve sketching, optimization problems and related rates. Definitions and theorems presented in this chapter are taken from [1] Larson, R. (2010), [2] Leithold, L. (2002) and [3] Stewart, J. (2016) At the end of this chapter, the student might be able to: 1. Apply the concept of differential in error propagation and in approximation. 2. Apply differentiation to determine the equations of tangent and normal lines to a graph of function at a given point 3. Apply the concepts of Relative Extrema, First Derivative Test, Second Derivative Test, Concavity and Points of Inflection in drawing a sketch of the graph of algebraic and transcendental functions 4. Apply the concepts of derivatives in solving optimization problems and in related rates problems 4.1. The Differential Consider a function defined by y=f(x) where x is the independent variable. In the three-step rule we introduced the symbol Δx to denote the increment of x. Now we introduce the symbol dx which we call the differential of x. Similarly, we shall call the symbol dy as the differential of y. To give separate meanings to dx and dy, we shall adopt the following definitions of a function defined by the equation y=f(x) Definition4.1.1.a Differential of the Dependent Variable If the function f is defined by the equation y  f x , then the differential of y, denoted by dy, is given by dy  f '(x)x where x is in the domain of f ' and x is an arbitrary increment of x. 127

MATH 401 – DIFFERENTIAL CALCULUS In words, the differential of a function is equal to its derivative multiplied by the differential of its independent variable. Definition4.1.1.b Differential of the Independent Variable If the function f is defined by the equation y  f x , then the differential of x, denoted by dx, is given by dx  x where x is any number in the domain of f ' and x is an arbitrary increment of x. In words, the differential of the independent variable is equal to the increment of the variable From Definitions 4.1 and 4.2, dy  f '(x)dx We emphasize that the differential dx is also an independent variable; it may be assigned any value whatsoever. Therefore, from DEFINITION 4.1, we see that the differential dy is a function of two independent variables x and dx. It should also be noted that while dx = Δx, dy ≠ Δy in general. Suppose dx≠0 and we divide both sides of the equation dy  f '(x)dx by dx then we get dy  f 'x . Note that this time dy/dx denotes the quotient of two dx differentials, dy and dx . Thus the definition of the differential makes it possible to define the derivative of the function as the ratio of two differentials. That is, f 'x  dy  the differential of y dx the differential of x Example1. Given f (x)  x2  x 1, find dy Solution: f 'x  2x 1 0 f 'x  2x 1 dy  f 'xdx dy  2x 1dx Example2. Find the differential of y  x3  5x 1 Solution: y  f x 128

MATH 401 – DIFFERENTIAL CALCULUS f x  x3  5x 1 f 'x  3x2  51 0 f 'x  3x2  5 dy  f 'xdx  dy  3x2  5 dx Example3. Find the differential of f (x)  x2 x4 1 Solution: f (x)  x2 x4 1 f '(x)  x 2  4x3 0  x4 12x   2 x4 1  4x5  x4 1   2x5  x4 1 f '(x)     2x    2x 2 x4 1  x4 1  f '(x)  2x5  2x x4 1  2x5  2x5  2x x4 1 x4 1 f '(x)  4x5  2x x4 1 dy  f 'xdx dy   4 x5  2x dx x 4 1 dy Example4. Find by means of differentials if xy + sin x = ln y dx Solution: x dy  y dx  cos x dx  1 dy y  x dy  y dx  cos x dx  1 dy y y xy dy  y2dx  y cos x dx  dy xy dy  y2dx  y cos x dx  dy 129

MATH 401 – DIFFERENTIAL CALCULUS  xy dy  y2dx  y cos x dx  dy 1 dx xy dy  y2  y cos x  dy dx dx xy dy  dy   y2  y cos x dx dx xy 1 dy   yy  cos x dx dy   yy  cos x dx xy 1 4.1.1. Application of the Differential Derivatives can also be used in finding the change in a certain quantity. For instance in the function y  f x, the change in x and y are denoted by x and y while dx and dy are the differentials respectively. Example1. Use differentials to approximate the change in the area of a square if the length of its side increases from 6 cm to 6.23 cm Solution: Let x = length of the side of the square. The area may be expressed as a function of x, where A = x2. The differential dA is dA  f 'x dx  dA  2x  dx Because x is increasing from 6 to 6.23, you find that Δ x = dx =0.23cm hence, dA  26cm0.23cm dA  2.76cm2 The area of the square will increase by approximately 2.76cm2 as its side length increases from 6 to 6.23. Note that the exact increase in area y is 2.8129cm2. Example2. Use the local linear approximation to estimate the value of 3 26.55 to the nearest thousandth. Solution: Because the function we are applying is f x  3 x , choose a convenient value of x that is a perfect cube and is relatively close to 26.55, namely x  27 . The differential dy is 130

MATH 401 – DIFFERENTIAL CALCULUS dy  f 'xdx 1 f x  x3 f 'x  1 x 1 1  f 'x  1 x  2  f 'x  1 3 3 3x2 3 3 3 dy  1 3 dx 3x2 Because x is decreasing from 27 to 26.55, then x  dx  0.45 Hence, dy   1 3  0.45   319  45    1   45    45  1  0.0167   100   27  100  2700 60  3272  which implies that 3 26.55 will be approximately 1 less that 3 27  3 60 therefore, 3 26.55  3  1 60 3 26.55  3  0.0167 3 26.55  2.9833 Example3. Use an appropriate local linear approximation to estimate the value of cos 310 Solution: Let y  cos x then dy   sin xdx  When x  30 , y  cos x  cos 30  0.8660  And when     0.01745 , then x  30 , x  dx  1  1  180   dy  sin xdx  sin30 0.01745   0.50.01745  0.008725 Therefore the required approximation is y  dy  0.8660   0.008725 y  dy  0.8573 131

MATH 401 – DIFFERENTIAL CALCULUS 4.1.2. Approximate Formulas The discussion on the geometric interpretation of the derivative f or y is given as follows; f  f x  x  f x f x  x  f x f In the given figure the comparison of f and df will be shown From the figure, it can be seen that dx  x but dy  y . However if x is very small or when x  0 then dy will be approximately equal to y . That is in symbols dy  y or equivalently df  f . Imposing the relation df  f to equation f x  x  f x f we will have, f x  x  f x  df . Since df  f ' (x)dx the relation will finally given as f x  x  f x f '(x)dx Which is called the approximation formula Example1. Given f x  3x2  x 1, compute the value of f and df if x  1 and x  0.01 Solution: f  f x  x  f x since f x  3x2  x 1then 132

MATH 401 – DIFFERENTIAL CALCULUS f  [3(x  x)2  (x  x) 1] [3x2  x 1] f  {3[x2  2xx  (x)2 ]  (x  x) 1} [3x2  x 1] f  3x2  6xx  3(x)2  x  x 1 3x2  x 1 f  6xx  3(x)2  x since x  1 and x  0.01 then f  6(1)(0.01)  3(0.01)2  0.01 f  0.06  0.0003  0.01 f  0.0503 df  f '(x)dx since f x  3x2  x 1 , then f '(x)  3(2x) 1 0 f '(x)  6x 1 df  f '(x)dx df  (6x 1)dx since dx  0.01 , then df  [6(1) 1][0.01] df  [5][0.01] df  0.05 Example2. Given that f x  x compute df and f when x  2 1 x and x  0.01 Solution: df  f '(x)dx since f x  x , then 1 x f ' ( x)  (1  x)(1)  (x)(0  1)  1 x  x (1 x)2 (1 x)2 f ' ( x)  1 (1  x)2 df  f '(x)dx df  1 dx since x  dx  0.01 , then (1 x)2 133

MATH 401 – DIFFERENTIAL CALCULUS df  (1 1 2 [0.01]  2) df  0.01 9 df  0.001111 f  f x  x  f x since f x  1 x x then  f  (x  x)  x (1 x  x) (1 x) f  (x  x)(1 x)  (x)(1 x  x) (1 x  x)(1 x) f  x  x2  x  xx  x  x2  xx (1 x  x)(1 x) f  x (1 x  x)(1 x) since x  2 and x  0.01 then f  0.01  0.01 (1 2  0.01)(1 2) (3.01)(3) f  0.01 9.03 f  0.001107 Example3. Find the approximate value of 4 82 using the differential. Solution: Let y  f (x)  4 x Think of the nearest number to the given 82 which is a perfect power of the index number 4, 81 is the nearest number to 82 which is a perfect power of 4. Set x  81and used the approximation formula f x  x  f x f '(x)dx , then x  x  82 x  82  x , since x  81 then 134

MATH 401 – DIFFERENTIAL CALCULUS x  82  81 x  1 f (x)  4 x  (x)1 4 f ' ( x)  1 1 1  1 3 x4 x4 44 f '(x)  1 4(4 x )3 f x  x  f x f '(x)dx f x  x  4 x  1 dx 4(4 x )3 Since x  81 and x  dx  1 then f 81 1  4 81  1 (1) 4(4 81)3 f 82  3  1 4(3)3 f 82  3  1 108 f 82  3  0.009259 f 82  3.009259 4.1.3. Error Propagation We can also use differentials in Physics to estimate errors, say in physical measuring devices. In these problems, we’ll typically take a derivative, and use the “dx” or “dy” part of the derivative as the error. Then, to get percent error, we’ll divide the error by the total amount and multiply by 100. The other thing to remember is that when we are solving for an error, it can go either way, so we typically express our answers with a “±” Example 1. The volume of a cube is 125 in3. If the volume measurement is known to be correct to within 2.5 in3, estimate the error in the measurement of a side of the cube. 135

MATH 401 – DIFFERENTIAL CALCULUS Solution: We first write down what the necessary information given in the problem. We have V=125, and dV=2.5 ( “dx” is the error part of the equation). We want the error in the side of the cube, so we want ds. V  s3 dV  3s2ds Substitute and solve for ds. Note that since we know that V  s3 , V  125 , we know a side s  5 then dV  3s2ds 2.5  3(5)2 ds ds  2.5 75 ds  0.0333in The error in the measurement of a side of the cube is ±0.0333in. Example2. The radius of a sphere is measured to be 5 mm. If this measurement is correct to within 0.05 mm, a) Estimate the propagated error in the surface area of the sphere. b) Estimate the propagated error in the volume of the sphere. c) Estimate the percent error of the volume of the sphere. Solution: Given r = 5, and dr = 0.05 (“dx” is the error part of the equation). We need the error in the surface area (dA) and the error in the volume (dV). We have to remember the equations from Geometry. a) Given the surface area of the sphere, differentiate it with respect to r: A  4r 2 dA  8rdr Since r = 5, and dr = 0.05 and π = 3.1416, then dA  8rdr dA  8(3.1416)(5)(0.05) dA  6.2832mm2 The error in the measurement of the surface area of the sphere is ±6.2832mm2. b) Given the volume of the sphere, differentiate with respect to r: 136

MATH 401 – DIFFERENTIAL CALCULUS V  4 r3 3 dV  4  (3r 2 )dr 3 dV  4r 2dr Since r = 5, and dr = 0.05 and π = 3.1416, then dV  4(3.1416)(5)2 (0.05) dV  15.708mm3 The error in the measurement of the volume of the sphere is ±15.708mm3. c) To get percent error: Percent Error = ErrorVolume(100) = dV (100) = dV (100) V  4 r3  3  = 15.708 (100)  15.708 (100)  (0.03)(100)  4 ( )(5)3 523.6 3 = 3% The percent error in the measurement of the volume of the sphere is 3%. Exercise 1. Find the derivative of the following functions by applying the definition of derivatives. 1. Given the following, find a. df b. f a. y  x2  3x  4 b. y  4x2  3x  1 , x  2 and x  0.2 c. f x  1 , x  2 and x  0.02 x2 2. Find the approximate value of 143 using differential 3. Find the approximate value of 3 128 using differential 4. The measurement of an edge of a cube is 15cm. If it is to be painted with thickness of paint equal to 0.01cm. Find a) actual amount of paint; b) approximate amount of paint. 137

MATH 401 – DIFFERENTIAL CALCULUS 5. The measurement of the base and altitude of a triangle are 20 cm and 30 cm, respectively. The possible error in each measurement is 0.3 cm. a) Estimate the possible propagated error in computing the area of the triangle. b) Approximate the percent error in computing this area. 4.2. Tangent Line and Normal Line to a curve (Algebraic and Transcendental Functions) The geometric interpretation of the derivative showed that the derivative of a function y  f (x) at any point P(x, y) is equal to the slope of the tangent line TL at the point P(x, y) . That is f '(x) at P(x, y) is equal to the slope of TL at P(x, y) . For a particular point P0(x0, y0) , the slope of the  tangent line TL will be f '(x0) . Knowing a point, P0(x0, y0) and the slope, f '(x0) of the tangent line we can use the point-slope form of the line to get the equation of the tangent line as y  y0  f '(x0)(x  x0) Similarly, we can get the equation of the nomal line NL to the same curve at the same point. Since the normal line is perpendicular to the tangent line, the slope of the normal line is equal to the negative reciprocal of the slope of the tangent line. So, if the slope of TL is f '(x0) then the slope of NL will be  1 , hence, the equation NL will be f '(x0 ) y  y0   f 1 (x  x0 ) . '(x0 ) Example1. Find the equations of TL and NL to the curve y  3x2  2x 1 at (2,9) Solution: Given y  f x  3x2  2x 1, x0  2 and y0  9 We first compute the slope of the tangent line f x  3x2  2x 1 f 'x  6x  2 f '2  6(2)  2  12  2 138

MATH 401 – DIFFERENTIAL CALCULUS f '2  10 Slope of Tangent Line Equation of Tangent Line (TL) Equation of Normal Line (NL) y  y0  f 'x0 x  x0  y  y0  f 1  x  x0  y  9  10x  2 ' x0 y  9  10x  20 y  9  1x  2 10x  20  y  9  0 10x  y 11  0 10 10y  9  x  2 10y  90  x  2 x 10y  92  0 Example2. Find the equations of T and N given the implicit equation of the curve (x  3y)2  8x 12  0 at point (2,0) . Solution: Given (x  3y)2 8x 12  0 , x0  2 and y0  0 We first compute the slope of the tangent line using implicit differentiation 2(x  3y)(1 3y')  8  0  0 2(x  3xy'3y  9yy')  8  0 2x  6xy'6y 18yy'8  0 (6xy'18yy')  (2x  6y 8) y'(6x 18y)  (2x  6y 8) y' (2x  6y  8) then (6x 18y) y' f 'x  (2x  6y  8) (6x 18y) since x  2 then f ' 2  (2)(2)  (6)(0)  8 (6)(2)  (18)(0) f ' 2  4  0  8 12  0 f ' 2   4 12 139

MATH 401 – DIFFERENTIAL CALCULUS Equation of Normal Line (NL) f ' 2  1 Slope of Tangent Line y  y0  f 1  x  x0  3 ' x0 Equation of Tangent Line (TL) y  0  1 x   2 1 y  y0  f 'x0 x  x0  3 y  0   1x   2 y  3x  2 3 y  3x  6 3y  (x  2) 3y  x  2) 3x  y  6  0 x  3y  2  0 Example3. Find the equation of the tangent line to the curve y2  8x  4 which is perpendicular to the line x  2y  5. Solution: Let mTL as slope of Tangent Line and mGL as slope of Given Line, then mTL  1 mGL from the equation of line x  2y  5 compute the slope of the given line. Using y  mx  b then 2y  x 5 y  1x  5 therefore mGL  1 22 2 mTL  1  1 2 then mTL  2 , Slope of Tangent Line mGL 1 2 For y2  8x  4 , compute y’ using implicit differentiation 2 yy, 8  y,  8  y,  4 since mTL  y' , then 2  4 2y y y 140

MATH 401 – DIFFERENTIAL CALCULUS y2 Substitute the value of y  2 to the equation y2  8x  4 and solve for x y2  8x  4  (2)2  8x  4  8x  4  4  8x  8 x 1 Therefore the value of (x0 , y0 ) is 1,2 Equation of Tangent Line (TL) y  y0  f 'x0 x  x0  y  2  2x 1 y  2  2x  2 2x  2  y  2  0 2x  y  0 Example4. Find the equations of a Tangent and Normal line in a given curve to the given value of x a. y  tan1 x , x = 1 Solution: Solve for the point of tangency y  tan1 x since x = 1 y  tan11 y 4 value of (x0 , y0 ) is 1,     4 Find the Slope f x  y  tan1 x f ' x  1 1 2 x f '1  1 1 11 1 12 141

MATH 401 – DIFFERENTIAL CALCULUS f '1  1 slope of tangent line 2 Equation of Tangent Line (TL) Equation of Normal Line (NL) y  y0  f 'x0 x  x0  y  y0  f 1  x  x0  ' x0 y     1 x 1 y    1 x 1 4 2 41 2 2 y     (x 1) y    2x 1  4 2y    x 1 4 2 y    2x  2 x  2y 1   0 4 2 2x  y  2    0 4 b. y  x2ex , x = 1 Solution: Solve for the point of tangency y  x2ex since x = 1 y  (1)2 e1 y  e1  value of (x0 , y0 ) is 1, e1 Find the Slope f x  y  x2ex f 'x  x2 (ex )  (ex )(2x) f 'x  x2ex  2xex since x = 1, then f '1  (1)2 e1  2(1)e1 f '1  e1  2e1 142

MATH 401 – DIFFERENTIAL CALCULUS Equation of Normal Line (NL) f '1  e1 slope of tangent line y  y0  f 1  x  x0  Equation of Tangent Line (TL) ' x0 y  y0  f 'x0 x  x0  y  e1  e1(x 1) y  e1  1 x  1 e1 y  e1  xe1  e1 xe1  e1  y  e1  0 e1( y  e1)  x 1 xe1  y  0 ye1  e2  x  1 x  ye1  e2 1  0 c. y  ln(x  3)4 @ x = -2 Solution: Solve for the point of tangency y  ln(x  3)4 since x = -2 y  ln(2  3)4 y  ln(1)4 y  ln1 y0 value of (x0 , y0 ) is  2,0 Find the Slope f x  y  ln(x  3)4 f x  4ln(x  3) f ' x  4 x 1 3(1)  f 'x  4 , since x = -2 , then x3 f ' 2  4  4 23 1 143

MATH 401 – DIFFERENTIAL CALCULUS f ' 2  4 slope of tangent line Equation of Tangent Line (TL) Equation of Normal Line (NL) y  y0  f 'x0 x  x0  y  y0  f 1  x  x0  y  0  4x  (2) ' x0 y  4x  2 y  0  1x   2 y  4x  8 4x  y  8  0 4 4y  x  2 4y  x  2 x  4y  2  0 d. y  sinx @ x  1 6 Solution: Solve for the point of tangency y  sinx since x  1 6 y  sin  1 6 y  sin     sin(30 ) 6 y  1 2 value of (x0 , y0 ) is  1, 1   6 2 Find the Slope f x  y  sin x f 'x  cosx  f ' x   cos x since x  1 , then 6  f ' x   cos   1   cos      cos  30    3  , 6 6 2 f ' x   3 slope of tangent line 2 144

MATH 401 – DIFFERENTIAL CALCULUS Equation of Tangent Line (TL) Equation of Normal Line (NL) y  y0  f 'x0 x  x0  y  y0  f 1  x  x0  ' x0 y    1    3  x   61  y    1   1 x   1   2    2 3  6 2 2 y  1   3 x  1   3  y  1    x  1  2 6  2  2  6 2 y1  3x 3  3 y   3  x  1 22 12 24 6  3xy 3 1 0 x 3 y 3 1 0 2 12 2 2 46 e.  f x  xcoshx @ 1,1 Solution: Find the Slope Let y  f x  xcosh x y  xcosh x Apply ln to both sides ln y  ln xcosh x ln y  cosh x(ln x) Differentiate implicitly with respect to x. 1  dy   cosh x 1   ln xsinh x y  dx   x Solve dy at 1,1 dx 1  dy   cosh11  ln1sinh1 1  dx  1 dy  cosh1  0sinh1 dx dy  cosh1 dx f ' x  dy dx 145

MATH 401 – DIFFERENTIAL CALCULUS f ' x  cosh1 slope of tangent line Equation of Normal Line (NL) Equation of Tangent Line (TL) y  y0  f 1  x  x0  y  y0  f 'x0 x  x0  ' x0 y 1  cosh1(x 1) y 1  1 (x 1) y 1  xcosh1 cosh1 cosh1 xcosh1 y  cosh11  0 (cosh1)(y 1)  1(x 1) y cosh1 cosh1  x 1 x  y cosh1 cosh11  0 Exercise 2. Find the equation of Tangent Line and Normal Line of the following curves at the given point. 1. y  x3  3x2  2 at x0 1. 2. y  3x2  2x 1 at (1,2) . 3. x2  xy  2y  2  0 at x  2 . 4. Find the equation of the tangent line to the curve x2  4y2  8 and parallel to the line x  2y  8 . 5. Find the equation of the tangent line to the curve y  x4 14x2 17x  40 and perpendicular to the line x  7 y  4 . 6. y  arcsin 2x , x = ¼ 7. y  log x , x = e 8. y  sec1 x , x = -2 9. y  ex ln2  x, x = 0 10. y  e2x 1 . x  1 2 146

MATH 401 – DIFFERENTIAL CALCULUS 4.3. Relative Extrema An important application of the derivative is to determine where a function attains its maximum and minimum (extreme) values. Definition4.3.1.a Relative Maximum Value The Function f has a relative maximum value at the number c if there exists an open interval containing c, on which f is defined, such that f(c) ≥ f(x) for all x in this interval. Figure below show a portion of the graph of a function having a relative maximum value at c. acb x acb x Figure 1 Figure 2 Definition4.3.1.b Relative Minimum Value The Function f has a relative minimum value at the number c if there exists an open interval containing c, on which f is defined, such that f(c) ≤ f(x) for all x in this interval. Figure below show a portion of the graph of a function having a relative minimum value at c. acb x acb x 147 Figure 3 Figure 4