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

**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: \[ f'(x)=12.9(\ln x)^2\cdot\frac{1}{x}=\frac{12.9(\ln x)^2}{x}. \]