Chapter 12: Additional Differentiation Topics PDF

MAA00B1 CHAPTER 12 Additional Differentiation Topics Chapter Objectives To develop a differentiation formula for y = In u, to apply the formula, and to use it to differentiate a logarithmic function to a base other than e. To develop a differentiation formula for y = eu , to appl...

MAA00B1 CHAPTER 12 Additional Differentiation Topics Chapter Objectives To develop a differentiation formula for y = In u, to apply the formula, and to use it to differentiate a logarithmic function to a base other than e. To develop a differentiation formula for y = eu , to apply the formula, and to use it to differentiate an exponential function with base other than e. To give a mathematical analysis of the economic concept of elasticity. To discuss the notion of a function defined implicitly and to determine derivatives by means of implicit differentiation. To describe the method of logarithmic differentiation and to show how to differentiate a function of the form u v. To find higher-order derivatives both directly and implicitly.  Chapter Objectives 12.1 Derivatives of Logarithmic Functions 12.2 Derivatives of Exponential Functions 12.3 Elasticity of Demand 12.4 Implicit Differentiation 12.5 Logarithmic Differentiation 12.7 Higher-Order Derivatives 12.1 Derivatives of Logarithmic Functions BASIC RULE 3: Derivative of ln x d 1 ( ln x ) = for x  0 dx x d dx ( ln x ) = for x  0 1 x Let y = ln u , where u is a differentiable function of x. d dx ( ln u ) =  1 du u dx for u  0 Example 1 – Differentiating Functions Involving ln x a. Differentiate f ( x) = 5ln x. d 1 5 Solution: We have f '( x) = 5 (ln x) = 5  = for x  0 dx x x ln x b. Differentiate y= 2. x Solution: By the quotient rule and Basic Rule 3, 1 2 x d ( ) ( ) x ln x − ln x ( ) d 2 x 2   − (ln x) ( 2 x )  x x − 2 x ln x 1 − 2ln x y ' = dx dx = = = for x  0 ( ) 2 4 4 3 x 2 x x x Example 3 – Rewriting Logarithmic Functions before Differentiating dy a. Find if y = ln(2 x + 5) 3. dx Solution: We have y = ln ( 2 x + 5 ) = 3ln ( 2 x + 5 ) for 2 x + 5  0. 3 dy  1  6 = 3 ( ) 2 = for x  −5 / 2 dx  2x + 5  2x + 5 b. Find f ( p) if f ( p) = ln (( p + 1) 2 ( p + 2)3 ( p + 3) 4 ). Solution: We simplify the right side, then differentiate. f ( p ) = 2ln( p + 1) + 3ln( p + 2) + 4ln( p + 3)  1   1   1  2 3 4 ( p) = 2  f′  (1) + 3   (1) + 4   (1) = + +  p +1  p+2  p +3 p +1 p + 2 p + 3  The change-of-base formula allows us to write: ln u y = logb u = ln b Example 5 – Differentiating a Logarithmic Function to the Base 2 Differentiate y = log 2 X. dy d  ln x  Solution: ( log 2 x ) = dx  ln 2  dx   1 d 1 log 2 e = (ln x) = , which can be expressed as. ln 2 dx ( ln 2 ) x x d logb e du More generally, (logb u) = . dx u dx 12.2 Derivatives of Exponential Functions The following is the Inverse Function Rule: d −1 1 −1 ( f ( x)) = −1 for f ′ ( f ( x))  0 dx f′ ( f ( x)) It allows us to differentiate exponential function when dy 1 rewritten as = dx dx dy Consequently, for u a differentiable function of x, we have d u u du (e ) = e dx dx Example 1 – Differentiating Functions Involving ex d a. Find (3e x ). dx d d x Solution: (3e ) = 3 e = 3e x x dx dx x dy b. If y= x , find. e dx Solution: y = xe , so dy d −x −x = ( x )e + x dx dx d −x dx e ( ) −x −x −x 1− x = (1)e + x(e )(−1) = e (1 − x) = x e Example 5 – Differentiating Different Forms Find d 2 dx ( e + xe + 2 x. ) Solution: d 2 dx ( e x e −1 e + x + 2 = 0 + ex + ) d  (ln 2) x  dx  e   ( ln 2 )  1  2 x ln 2 = ex e −1 + e ( ln 2 ) x   = ex e −1 +   2 x  2 x Example 6 – Differentiating Power Functions Again For x  0 , we can write x =ea a ln x. So we have d a dx x = ( ) d a ln x dx e = e a ln x d dx ( a ln x ) = x a ax −1 = ax ( a −1. ) Note: here we do not require that a be a positive integer. 12.3 Elasticity of Demand If p = f (q ) is a differentiable demand function, the point elasticity of demand, denoted by  ( eta ) , at (q, p) is given by p  = (q) = q dp dq There are three categories of elasticity: 1. When |  |  1 , demand is elastic. 2. When |  | = 1 , demand has unit elasticity. 3. When |  |  1 , demand is inelastic. Example 1 – Finding Point Elasticity of Demand Determine the point elasticity of the demand equation k p = , where k  0 and q  0. q Solution: From the definition, we have p k = q dp = q2 −k = −1 dq q2 Thus, the demand equation has unit elasticity for all q  0. Another useful expression for η is when q = g ( p) : g′( p)  =  ( p) =. g ( p) 12.4 Implicit Differentiation Implicit Differentiation Procedure For an equation that we assume defines y implicitly as a dy differentiable function of x, the derivative can be found dx as follows: 1. Differentiate both sides of the equation with respect to x. dy 2. Solve for , nothing any restrictions. dx  Example 1 – Implicit Differentiation Find dy by implicit differentiation if y + y 3 – x = 7. dx Solution: Differentiating both sides with respect to x, d dx ( d ) y + y − x = (7) 3 dx dy 2 dy dy dy + 3y − 1 = 0 Solving for gives (1 + 3 y 2 ) = 1 dx dx dx dx dy 1 =. Note that there is no restriction on y. dx 1 + 3 y 2 Example 3 – Implicit Differentiation Find the slope of the curve x 3 = ( y – x 2 2 ) at (1, 2). dy Solution: Finding by implicit differentiation, we have dx d 3 ( ) d  y−x  ( ) 2 x = 2 dy 3x 2 + 4 xy − 4 x 3 dx   = − 0 2 for y x dx dx ( 2 y−x 2 )  d  2 (  dx 2 ) 3x = 2 y − x  ( y − x 2 )   dy 2( y − x 2 ) = 3x 2 + 4 xy − 4 x 3  dy  dx 2 ( 3x = 2 y − x  − 2 x   dx  2 ) dy 7 7 =. Thus, the slope of the curve at (1, 2) is. dx (1,2) 2 2 dy dy 3x 2 = 2 y − 4 xy − 2 x 2 + 4 x3 dx dx 12.5 Logarithmic Differentiation Logarithmic Differentiation Procedure To differentiate y = f ( x), 1. Take the natural logarithm if both sides. This results in ln y = ln ( f ( x)) 2. Simplify ln ( f ( x)) by using properties of logarithms. 3. Differentiate both sides with respect to x. 4. Solve for dy. dx 5. Express the answer in terms of x only. This requires substituting f ( x) for y. Example 1 – Logarithmic Differentiation ( 2 x − 5) 3 Find y if y=. x 24 x +1 2 ( 2 x − 5) 3 Solution: Using the steps listed above, we have ln y = ln x2 4 x2 + 1 ln y = ln ( 2 x − 5 ) − (ln x 2 + ln( x 2 + 1)1/4 ) 3 1 = 3ln ( 2 x − 5 ) − 2ln x − ln( x 2 + 1). Differentiating with respect to x, 4 y′  1  1 1 1  = 6 − 2 − x = 3  (2) − 2   −  2  (2 x). Solving for y′yields y  2x + 5   x  4  x +1 2 x − 5 x 2( x + 1) 2  ( 2 x − 5)  6 3  6 2 x 2 x  y′= y  − − = 24 2  − − .  2 x − 5 x 2( x + 1)  x x + 1  2 x − 5 x 2( x + 1)  2 2 Example 3 – Relative Rate of Change of a Product Show that the relative rate of change of a product is the sum of the relative rates of change of its factors. Use this result to express the percentage rate of change in revenue in terms of the percentage rate of change in price. Solution: Suppose r is a function of an unspecified variable x. We are to show that if r = pq where also p and q are functions of x, then r ′ p′ q′ = +. r p q Example 3 – Continued Solution, continued From r = pq we have ln r = ln p + ln q. r ′ p′ q′ Differentiating with respect to x gives = + as required r p q r′ p′ q′ Multiplying by 100% give 100% = 100% + 100%. r p q From the elasticity of demand section, and specifying our variable x to be p, we can write r′ p′ 100% = (1 +  ) 100%, r p expressing the percentage rate of change in revenue in terms of the percentage rate of change in price. 12.7 Higher-Order Derivatives Some notations for higher-order derivatives are given in the table below: Table 12.3 y super dash baseline. f super dash baseline left parenthesis x right parenthesis. d y over dx. d over dx times left parenthesis f function of x right parenthesis. D sub x baseline y. First ' d y d derivative: y' f ( x) dx dx ( f ( x)) Dx y d2 y super dash dash baseline. f super dash dash baseline left parenthesis x right parenthesis. d super 2 baseline y over dx super 2 baseline. d super 2 baseline over dx super 2 baseline times left parenthesis f function of x right parenthesis. D sub x baseline super 2 baseline y. Second d2y Dx2 y derivative: y '' f '' ( x) dx 2 2 ( f ( x)) dx d3y y super dash dash dash baseline. f super dash dash dash baseline left parenthesis x right parenthesis. d super 3 baseline y over dx super 3 baseline. d super 3 baseline over dx super 3 baseline times left parenthesis f function of x right parenthesis. D sub x baseline super 3 baseline y. 3 Third d derivative: y ''' f ''' ( x) ( f ( x)) Dx3 y dx3 3 dx d4y y super left parenthesis 4 right parenthesis baseline. f super left parenthesis 4 right parenthesis baseline left parenthesis x right parenthesis. d super 4 baseline y over dx super 4 baseline. d super 4 baseline over dx super 4 baseline times left parenthesis f function of x right parenthesis. D sub x baseline super 4 baseline y. Forth y (4) f (4) ( x) d 4 Dx4 y derivative: dx 4 ( f ( x)) dx 4 y super left parenthesis n right parenthesis baseline. f super left parenthesis n right parenthesis baseline left parenthesis x right parenthesis. n d super n baseline y over dx super n baseline. d super n baseline over dx super n baseline times left parenthesis f function of x right parenthesis. D sub x baseline super n baseline y. nth d y n Derivative: y (n) f (n) ( x) d ( f ( x)) Dxn y dx n dx n Example 1 – Finding Higher-Order Derivatives If f ( x) = 6 x3 − 12 x 2 + 6 x − 2, find all higher-order derivatives. Solution: Differentiating gives f ( x) = 18 x 2 − 24 x + 6 Differentiating f ( x) gives f ( x) = 36 x – 24 Similarly, f  ( x ) = 36 f( 4) ( ) x = 0, and for n  5, f ( n) ( x ) = 0. Example 3 – Evaluating a Second-Order Derivative 16 d2y If y = f ( x) = , find and evaluate it when x = 4. x+4 dx 2 Solution: Since y = 16( x + 4)−1 , dy = −16 ( x + 4 ) −2 the power rule gives dx d2y 32 = 32 ( x + 4 ) = −3 and differentiating again,. dx 2 ( x + 4) 3 d2y Evaluating when x = 4, we obtain = 32 1 =. dx 2 x=4 83 16 1 Thus, y ''(4) =. 16 Example 5 – Higher-Order Implicit Differentiation d2y Find 2 if x 2 + 4 y 2 = 4. dx dy Solution: Differentiating, we obtain 2x + 8 y =0 dx dy − x =. Differentiating again using the quotient rule, dx 4 y d d  dy  2 4 y (− x) − (− x) (4 y ) 4 y (−1) − (− x)  4  d y = dx dx =  dx  dx 2 (4 y ) 2 16 y 2 dy dy −4 y + 4 x −y + x = dx = dx. 16 y 2 4 y2 Example 5 – Continued Solution, continued It is customary to express the answer without the derivative, in terms of x and y only:  −x  −y + x  2 d y  = 4 y −4 y 2 − x 2 4 y 2 + x 2 2 = 2 3 =− 3. dx 4y 16 y 16 y 2 d y 1 Finally, since x + 4 y = 4, 2 2 2 =− 3. dx 4y