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d tant lea) de tan' L) Itx? 2x THlege THe" ux 2) la 2x ln x Ids 2 lox t 2 y de 4(2/ax+2) 3) fL7)-6 gi(7)=4 (x=7) t f(G) し -50 t 24 -26

d tant lea) de tan' L) Itx? 2x THlege THe" ux 2) la 2x ln x Ids 2 lox t 2 y de 4(2/ax+2) 3) fL7)-6 gi(7)=4 (x=7) t f(G) し -50 t 24 -26

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.2: Derivatives Of Products And Quotients
Problem 35E
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### Calculus Homework Solutions The following notes cover derivative problem solutions typically encountered in a Calculus course. --- #### Problem 1 Evaluate the derivative of the function \( \tan^{-1}(e^x) \): \[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] \] We know that: \[ \frac{d}{dx} \left[ \tan^{-1}(x) \right] = \frac{1}{1 + x^2} \] Using the chain rule: \[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] = \frac{1}{1 + (e^x)^2} \cdot \frac{d}{dx}(e^x) \] Simplify the expression: \[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] = \frac{1}{1 + e^{2x}} \cdot e^x = \frac{e^x}{1 + e^{2x}} = \frac{2e^{2x}}{1 + e^{2x}} \] --- #### Problem 2 Given the function \( y = 2x \ln x \), find \( \frac{dy}{dx} \): Use the product rule: \[ y = 2x \ln x \] \[ \frac{dy}{dx} = 2 (\ln x) + 2x \left( \frac{1}{x} \right) \] Simplify the expression: \[ \frac{dy}{dx} = 2 (\ln x) + 2 \] Express \( \frac{dy}{dx} \) in terms of \( y \): \[ \frac{dy}{dx} = y (2 \ln x + 2) \] --- #### Problem 3 Given the functions and their derivatives: \[ f(7) = 6, \quad f'(7) = 10 \] \[ g(7) = -5, \quad g'(7) = 4 \] Find \( h'(x) \) for \( h(x) = f(x) g(x) \) at \( x = 7 \): Using the product rule: \[ h'(x) = f'(x) g(x) +
Transcribed Image Text:### Calculus Homework Solutions The following notes cover derivative problem solutions typically encountered in a Calculus course. --- #### Problem 1 Evaluate the derivative of the function \( \tan^{-1}(e^x) \): \[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] \] We know that: \[ \frac{d}{dx} \left[ \tan^{-1}(x) \right] = \frac{1}{1 + x^2} \] Using the chain rule: \[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] = \frac{1}{1 + (e^x)^2} \cdot \frac{d}{dx}(e^x) \] Simplify the expression: \[ \frac{d}{dx} \left[ \tan^{-1} (e^x) \right] = \frac{1}{1 + e^{2x}} \cdot e^x = \frac{e^x}{1 + e^{2x}} = \frac{2e^{2x}}{1 + e^{2x}} \] --- #### Problem 2 Given the function \( y = 2x \ln x \), find \( \frac{dy}{dx} \): Use the product rule: \[ y = 2x \ln x \] \[ \frac{dy}{dx} = 2 (\ln x) + 2x \left( \frac{1}{x} \right) \] Simplify the expression: \[ \frac{dy}{dx} = 2 (\ln x) + 2 \] Express \( \frac{dy}{dx} \) in terms of \( y \): \[ \frac{dy}{dx} = y (2 \ln x + 2) \] --- #### Problem 3 Given the functions and their derivatives: \[ f(7) = 6, \quad f'(7) = 10 \] \[ g(7) = -5, \quad g'(7) = 4 \] Find \( h'(x) \) for \( h(x) = f(x) g(x) \) at \( x = 7 \): Using the product rule: \[ h'(x) = f'(x) g(x) +
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ISBN:
9780321964038
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GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:
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