Section The Chain Rule and Implicit Differentiation with Application on Derivative of Logarithm Functions

Transcription

1 Section The Chain Rule an Implicit Differentiation with Application on Derivative of Logarithm Functions Ruipeng Shen September 3r, 5th Ruipeng Shen MATH 1ZA3 September 3r, 5th 1 / 3

2 The Chain Rule: Derivative of a Composite Function Assume y = f u) an u = g). Then y = F ) = f g) is a composite function. Theorem The Chain Rule) If g is ifferentiable at an f is ifferentiable at g), then the composite function F = f g efine by F ) = f g)) is ifferentiable at an F is given by the prouct F ) = f g)) g ) In Leibniz notation, if y = f u) an u = g) are both ifferentiable functions, then y = y u u Iea Let u = g). We have f g)) f ga)) a = f u) f ga)) u ga) g) ga). a a u = g) ga). As a result, we obtain f g) a) = f ga)) g a). Ruipeng Shen MATH 1ZA3 September 3r, 5th / 3

3 The First Eample Eample Fin the erivative of F ) = + 1. Solution 1 We can epress F ) = f g)) = f g)) where f u) = u an g) = + 1. Since f u) = 1 u an g ) =, we have F ) = f g)) g ) = = + 1. Solution If we let u = + 1 an y = u, then y = F ). Thus F ) = y = y u u = 1 u = + 1. Ruipeng Shen MATH 1ZA3 September 3r, 5th 3 / 3

4 Make the Calculation Easier: Combine Rules Let us consier the general case: If y is a composition of a power function y = f u) = u n an another function u = g), then y = y u u = nun 1 u. Formula The Power Rule Combine with the Chain Rule) If n is any number an u = g) is ifferentiable, then un n 1 u ) = nu g n ) = ng n 1 g Eample Fin the erivative of F ) = + + 1) 9. Solution F ) = ) ) = ) 8 + 1). Ruipeng Shen MATH 1ZA3 September 3r, 5th 4 / 3

5 More Combine Rules Formula Combine Rules) If f ) be a ifferentiable function, then ef ) ) = e f ) f ) [cos f )] = [ sin f )] f ) [sin f )] = [cos f )] f ) [tan f )] = [sec f )] f ) Eample Differentiate the function F ) = e Solution By the prouct rule, we have F ) = ) e + e ) = e + e ) = + 1)e. Ruipeng Shen MATH 1ZA3 September 3r, 5th 5 / 3

6 More Eamples Eample Differentiate the function G) = cos Solution By the chain rule an the quotient rule G ) = sin 1 ) ) = sin 1 ) 1 ) 1 + ) 1 ) 1 + ) ) = sin 1 ) 1 + ) 1 ) ) = sin 1 ) [ ] ) = 1 + ) sin Ruipeng Shen MATH 1ZA3 September 3r, 5th 6 / 3

7 More Eamples Eample Differentiate the function F ) = e sincos ). Solution 1 If y = e u, u = sin w, w = cos, then y = F ). Applying the chain rule twice, we have F ) = y = y u u = y u w u w. Thus F ) = e u )cos w) sin ) = e sincos ) [coscos )] sin. Solution By the combine rules, F ) = e sincos ) }{{} coscos ) }{{} sin ) }{{} Combine Rule Combine Rule Derivative of cos with Eponential with Sine F ) = e sincos) coscos ) sin. Ruipeng Shen MATH 1ZA3 September 3r, 5th 7 / 3

8 Application on the Derivative of Eponential Functions Formula If a > 0, then a ) = a ln a; af ) ) = a ln a)f ). Proof: We have a = e ln a ) = e ln a. By the chain rule, we obtain a ) = e ln a ) = e ln a ln a) = a ln a. Eample Differentiate the function f ) = tan. Solution By the combine rule, f ) = tan ln tan ) = ln ) tan sec. Ruipeng Shen MATH 1ZA3 September 3r, 5th 8 / 3

9 Implicit Differentiation: Eample Eample Consier the curve C efine by the equation 3 + y 3 = 6y. Fin the tangent line to C at the point P4/3, 8/3). 5.5 y=f ) 1 P A y =6y y=f ) y=f ) 3 =a -5 B Ruipeng Shen MATH 1ZA3 September 3r, 5th 9 / 3

10 Implicit Differentiation: Eample Continue) Let us assume y is a function of. Differentiating both sies of the equation 3 + y 3 = 6y, we obtain 3 + y 3 = 6y 3 + 3y y = 6 y + 6y 3 + 3y y = 6y + 6y 3y 6)y = 6y 3 y = y y 3 + [y)] 3 = 6[y)] 3 + 3[y)] y ) = 6y + 6 y ) y ) = y) [y)] Thus we can calculate the slope of the tangent line at P4/3, 8/3): slope = y 4/3) = 8/3) 4/3) 8/3) 4/3) = 4 5. As a result, we obtain the equation of the tangent line: y 8 3 = ). Ruipeng Shen MATH 1ZA3 September 3r, 5th 10 / 3

11 Implicit Differentiation: Eample Continue) Question Fin the point Aa, b) where the tangent line to C is vertical. The tangent line at Aa, b) is vertical. Thus the erivative y = y y is not well efine at A. This implies the enominator y is equal to zero at a, b). We have a, b) satisfies the equation { b a = 0; a 3 + b 3 = 6ab. Plugging a = b / into the secon equation, we have ) b 3 ) b + b 3 = 6 b b6 8 + b3 = 3b 3 b 6 = 16b 3 b 3 = 16. Thus b = 4/3 = a = b / = 8/3 1 = 5/3. Therefore A is 5/3, 4/3 ). Ruipeng Shen MATH 1ZA3 September 3r, 5th 11 / 3

12 Implicit Differentiation Proposition Let y be an implicit function of efine by an implicit equation. In orer to epress the erivative y/ in terms of an y, one can ifferentiate both sies of the equation with respect to an then solve the resulting equation for y/ or y ). Remark An implicit equation is usually given by F, y) = G, y). Eample Fin y an y if 4 + y 4 = r 4. Solution By implicit ifferentiation, we have 4 + y 4 = r 4 = y 3 y = 0 = y = 3 y 3. Ruipeng Shen MATH 1ZA3 September 3r, 5th 1 / 3

13 Eample: Fin the Secon Derivative Continue) In orer to epress f in term of an y, we can ifferentiate y an obtain y = y ) = 3 ) = 3 ) y 3 3 y 3 ) Plugging in y = 3 /y 3, we have the answer y 3 = 3 y 3 3 3y y y 6 = 3 y y y ) y 6. ] y 3 ) [ 3 y y 3 y = y 6 = 3 y 4 + 4) y 7 = 3r 4 y 7. In the last step, we use the equation 4 + y 4 = r 4. y 6 [ 3 y 3 y + 4 y = 3 y 7 ] Ruipeng Shen MATH 1ZA3 September 3r, 5th 13 / 3

14 Application: Derivative of Inverse Sine Function Recall the efinition of the inverse sine function arcsine function) y = sin 1 sin y = an π y π. Differentiating the equation sin y =, we obtain cos y) y = 1 y = = 1 cos y. By the ientity cos y = 1 sin y = 1 an the fact cos y 0, we have cos y = 1. Therefore Formula sin 1 ) = y = 1 cos y = , if 1 < < 1. 1 Ruipeng Shen MATH 1ZA3 September 3r, 5th 14 / 3

15 Application: Derivative of Inverse Tangent Function Recall the efinition of the inverse tangent function arctangent function) y = tan 1 tan y = an π < y < π. Differentiating the equation tan y =, we obtain sec y) y = 1 y = = 1 sec y. By the ientity sec y = 1 + tan y = 1 +. Therefore Formula tan 1 ) = y = 1 sec y = Ruipeng Shen MATH 1ZA3 September 3r, 5th 15 / 3

16 Derivatives of Inverse Trigonometric Function Formula Derivatives of inverse trigonometric functions are given by sin 1 ) 1 = csc 1 ) 1 = 1 1 cos 1 ) 1 = sec 1 ) 1 = 1 1 tan 1 ) = 1 cot ) = sin 1 f ) ) f ) = tan 1 f ) ) f ) = 1 [f )] 1 + [f )] Eample If F ) = tan 1 ), then F ) = ) ) = Ruipeng Shen MATH 1ZA3 September 3r, 5th 16 / 3

17 Derivative of an Inverse Function If f ) is a one-to-one ifferentiable function an its inverse function f 1 is also ifferentiable, then we can ifferentiate both sies of the ientity f f 1 )) = an obtain provie f f 1 )) 0. Theorem f f 1 )) f 1 ) ) = 1 = f 1 ) ) = 1 f f 1 )), If f ) is a one-to-one ifferentiable function an f f 1 a)) 0, then f 1 is ifferentiable at a an f 1 ) 1 a) = f f 1 a)). Ruipeng Shen MATH 1ZA3 September 3r, 5th 17 / 3

18 Derivative of Logarithm Functions If f ) = a then f 1 ) = log a. Since f ) = a ln a 0, we can apply the formula for erivative of the inverse function an obtain log a ) = f 1 ) 1 ) = f f 1 )) = 1 f log a ) = 1 a log a ln a = 1 ln a. Formula Derivative of Logarithm Functions) If a > 0 an a 1, then log a ) = 1 ln a. In particular, if a = e, we have ln ) = 1, ln u) = 1 u u, [ln g)] = g ) g). Ruipeng Shen MATH 1ZA3 September 3r, 5th 18 / 3

19 Eamples Eample Let f ) = ln. Fin the erivative f ). Solution I) If > 0, then f ) = ln. Therefore f ) = 1/. II) If < 0, then f ) = ln ). By the combine rule In summary, we have ln = 1. Eample f ) = 1 ) = 1. Differentiate the function f ) = lnsin ). Solution By the combine rule, we have f ) = sin ) sin = cos sin = cot. Ruipeng Shen MATH 1ZA3 September 3r, 5th 19 / 3

20 More Eamples Eample Differentiate the function g) = ln + + 1). Solution We can apply the chain rule an basic formulas: g ) = = = = = ) ) ) ) ln u) = 1 u u u) = 1 u u Ruipeng Shen MATH 1ZA3 September 3r, 5th 0 / 3

21 Eample of Logarithmic Differentiation Eample Differentiate the function y = Solution Take natural logarithms of both sies an ifferentiate ln y = ln ) = ln 1 y = y ) ln + ln ) 1 = ln + 1 = ln + y = y ln + = ln +. Alternative Metho We can also use the ientity y = = e ln ) = e ln an ifferentiate by the chain rule. Ruipeng Shen MATH 1ZA3 September 3r, 5th 1 / 3

22 Logarithmic Differentiation Proposition Steps in Logarithmic Differentiation) I) Take natural logarithms of both sies of an equation y = f ) an use the Laws of Logarithms to simplify. II) Differentiate implicitly with respect to. III) Solve the resulting equation for y. Eample Prove the formula n ) = n n 1 for an arbitrary real number n. Solution Let y = n. Applying the Logarithmic Differentiation, we obtain ln y = n ln ; 1 y y = n 1 ; y = y n = n n = n n 1. Ruipeng Shen MATH 1ZA3 September 3r, 5th / 3

23 Eample Eample Differentiate the function y =. + 1) Solution Applying the Logarithmic Differentiation, we obtain ln y = ln = ln ln + 1) ) [ + 1 = ln 1 ] ln + 1) 1 y = [ln 1 ] [ 1 y ln + 1) ) ] + 1 y = ln 1 ln + 1) + [ 1 = ln 1 ln + 1) = ) [ ln 1 ln + 1) ] ]. Ruipeng Shen MATH 1ZA3 September 3r, 5th 3 / 3