Question
1. (5 pts) Consider the following functions. State the inside and outside functions. Then find the derivatives with Form 2 of the chain rule. a. \( f(x)=\ln (5 x+11) \) \[ \begin{array}{l} \text { in rule. po not show the Form } 1 \text { steps. } \\ g(h)=5 \quad h(x)=5 \end{array} \] \[ f^{\prime}(x)=\frac{d f}{d x}=\frac{5}{x} \] \[ \begin{array}{ll} \text { b. } f(x)=3\left(17^{x^{5}}\right) & g(h)=3 \end{array} \quad h(x)=17^{x^{5}} \] c. \( f(x)=-2 e^{-x} \) \[ g(h)=-2 \quad h(x)=e^{-x} \] \[ f^{\prime}(x)=\frac{d f}{d x}=O\left(e^{-x}\right) \cdot e^{-x} \] 8 d. \( f(x)=-5 \sqrt[3]{d} \) \( f^{\prime}(x)=\frac{d f}{d x}= \) \[ \begin{array}{l} \text { e. } f(x)=4.3(\ln x)^{3}+e^{\pi} \quad g(h)=4.3(h)^{3+e^{h}} h(x)=\ln (x) \\ f^{\prime}(x)=\frac{d f}{d x}= \\ g^{\prime}(h(x)) \cdot h^{\prime}(x)=12.9(\ln x)^{2} \cdot \frac{1}{x}=\frac{12.9(\ln x)^{2}}{x} \end{array} \]
Ask by Summers Hodges. in the United States
Mar 12,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
**a. For \( f(x)=\ln(5x+11) \):**
- Inside function: \( h(x)=5x+11 \)
- Outside function: \( g(u)=\ln u \)
Derivative:
\[
f'(x)=\frac{5}{5x+11}
\]
**b. For \( f(x)=3\left(17^{x^5}\right) \):**
- Inside function: \( h(x)=17^{x^5} \)
- Outside function: \( g(u)=3u \)
Derivative:
\[
f'(x)=15x^4\ln 17\cdot 17^{x^5}
\]
**c. For \( f(x)=-2e^{-x} \):**
- Inside function: \( h(x)=e^{-x} \)
- Outside function: \( g(u)=-2u \)
Derivative:
\[
f'(x)=2e^{-x}
\]
**d. For \( f(x)=-5\sqrt[3]{x} \):**
- Inside function: \( h(x)=x \)
- Outside function: \( g(u)=-5u^{\frac{1}{3}} \)
Derivative:
\[
f'(x)=-\frac{5}{3}x^{-\frac{2}{3}}
\]
**e. For \( f(x)=4.3(\ln x)^3+e^{\pi} \):**
- Inside function: \( h(x)=\ln x \)
- Outside function: \( g(u)=4.3u^3+e^\pi \)
Derivative:
\[
f'(x)=\frac{12.9(\ln x)^2}{x}
\]
Solution

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The Deep Dive
The inside function in part (a) is \( h(x) = 5x + 11 \) and the outside function is \( g(h) = \ln(h) \). Using the chain rule, the derivative is \[ f^{\prime}(x) = \frac{1}{5x + 11} \cdot 5 = \frac{5}{5x + 11}. \] For part (b), the inside function is \( h(x) = 17^{x^5} \) and the outside function is \( g(h) = 3h \). The derivative is \[ f^{\prime}(x) = 3 \cdot \ln(17) \cdot 17^{x^5} \cdot 5x^4 = 15 \ln(17) \cdot x^4 \cdot 17^{x^5}. \]