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. $$ f(x)=-5 \sqrt[3]{d} $$ $$ f^{\prime}(x)=\frac{d f}{d x}= $$ $$ \begin{array}{l} \text { e. } f(x)=4.3(\ln x)^{3}+e^{\pi} \quad g(h)=4.3(h)^{3+e^{h}} h(x)=\ln (x) \ f^{\prime}(x)=\frac{d f}{d x}= \ g^{\prime}(h(x)) \cdot h^{\prime}(x)=12.9(\ln x)^{2} \cdot \frac{1}{x}=\frac{12.9(\ln x)^{2}}{x} \end{array} $$

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**a. For $$ f(x)=\ln(5x+11) $$**  
- Inside function: $$ h(x)=5x+11 $$  
- Outside function: $$ g(u)=\ln u $$

Differentiate using the chain rule (Form 2):  
$$
f'(x)=g'(h(x))\cdot h'(x)
$$
We have  
$$
g'(u)=\frac{1}{u} \quad \text{and} \quad h'(x)=5.
$$
Thus,  
$$
f'(x)=\frac{1}{5x+11}\cdot 5=\frac{5}{5x+11}.
$$

---

**b. For $$ f(x)=3\left(17^{\,x^5}\right) $$**  
- Inside function: $$ h(x)=17^{\,x^5} $$  
- Outside function: $$ g(u)=3u $$

Differentiate using the chain rule:  
$$
f'(x)=g'(h(x))\cdot h'(x).
$$
Since  
$$
g'(u)=3,
$$
we need $$ h'(x) $$. Write $$ h(x)=17^{x^5} $$ in exponential form:
$$
h(x)=e^{x^5\ln 17}.
$$
Differentiate:  
$$
h'(x)=e^{x^5\ln 17}\cdot \frac{d}{dx}(x^5\ln 17)=17^{x^5}\cdot 5x^4\ln 17.
$$
Thus,  
$$
f'(x)=3\cdot \left(17^{x^5}\cdot 5x^4\ln 17\right)=15x^4\ln 17\cdot 17^{x^5}.
$$

---

**c. For $$ f(x)=-2e^{-x} $$**  
- Inside function: $$ h(x)=e^{-x} $$  
- Outside function: $$ g(u)=-2u $$

Differentiate using the chain rule:  
$$
f'(x)=g'(h(x))\cdot h'(x).
$$
Since  
$$
g'(u)=-2,
$$
find $$ h'(x) $$:  
$$
h(x)=e^{-x}\quad \Rightarrow \quad h'(x)=-e^{-x}.
$$
Thus,  
$$
f'(x)=-2\cdot\left(-e^{-x}\right)=2e^{-x}.
$$

---

**d. For $$ f(x)=-5\sqrt[3]{x} $$**  
*(Assuming the variable under the cube root is $$ x $$ rather than $$ d $$.)*

- Rewrite the function as $$ f(x)=-5x^{\frac{1}{3}} $$.  
- Inside function: $$ h(x)=x $$  
- Outside function: $$ g(u)=-5u^{\frac{1}{3}} $$

Differentiate using the chain rule:  
$$
f'(x)=g'(h(x))\cdot h'(x).
$$
Here,  
$$
g'(u)=-5\cdot\frac{1}{3}u^{-\frac{2}{3}}=-\frac{5}{3}u^{-\frac{2}{3}},
$$
and  
$$
h'(x)=1.
$$
Thus,  
$$
f'(x)=-\frac{5}{3}x^{-\frac{2}{3}}.
$$

---

**e. For $$ f(x)=4.3(\ln x)^3+e^{\pi} $$**  
- Inside function: $$ h(x)=\ln x $$  
- Outside function: $$ g(u)=4.3u^3+e^\pi $$

Differentiate using the chain rule:  
$$
f'(x)=g'(h(x))\cdot h'(x).
$$
Here,  
$$
g'(u)=4.3\cdot3u^2=12.9u^2,
$$
and  
$$
h'(x)=\frac{1}{x}.
$$
Substitute back:  
$$
f'(x)=12.9(\ln x)^2\cdot\frac{1}{x}=\frac{12.9(\ln x)^2}{x}.
$$
**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: $$ f'(x)=\frac{12.9(\ln x)^2}{x} $$

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