Question

for the given composite function, identify the inner function, $u = g(x)$, and the outer function, $y = f(u)$. (use non - identity functions for $f(x)$ and $g(x)$.)
$y=sqrt3{1 + 7x}$
$(f(u),g(x))=(quad)$
find the derivative $\frac{dy}{dx}$.
$\frac{dy}{dx}=quad$
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1. -/2 points

write the composite function in the form $f(g(x))$. identify the inner function $u = g(x)$ and the outer function $y = f(u)$. (use non - identity functions for $f(u)$ and $g(x)$.)
$y = e^{sqrt{x}}$
$(f(u),g(x))=(quad)$
find the derivative $\frac{dy}{dx}$.

Explanation:

Step1: Identify inner and outer functions for $y = \sqrt[3]{1 + 7x}$

Let $u=g(x)=1 + 7x$ and $y = f(u)=\sqrt[3]{u}=u^{\frac{1}{3}}$. So $(f(u),g(x))=(u^{\frac{1}{3}},1 + 7x)$

Step2: Find derivatives of inner and outer functions

The derivative of $g(x)$ with respect to $x$ is $g^\prime(x)=\frac{d}{dx}(1 + 7x)=7$. The derivative of $f(u)$ with respect to $u$ is $f^\prime(u)=\frac{1}{3}u^{-\frac{2}{3}}$.

Step3: Apply chain - rule

By the chain - rule $\frac{dy}{dx}=f^\prime(g(x))\cdot g^\prime(x)$. Substitute $u = g(x)=1 + 7x$ into $f^\prime(u)$ and multiply by $g^\prime(x)$. We get $\frac{dy}{dx}=\frac{1}{3}(1 + 7x)^{-\frac{2}{3}}\cdot7=\frac{7}{3(1 + 7x)^{\frac{2}{3}}}$

Step4: Identify inner and outer functions for $y = e^{\sqrt{x}}$

Let $u = g(x)=\sqrt{x}=x^{\frac{1}{2}}$ and $y = f(u)=e^{u}$. So $(f(u),g(x))=(e^{u},x^{\frac{1}{2}})$

Step5: Find derivatives of inner and outer functions

The derivative of $g(x)$ with respect to $x$ is $g^\prime(x)=\frac{d}{dx}(x^{\frac{1}{2}})=\frac{1}{2}x^{-\frac{1}{2}}$. The derivative of $f(u)$ with respect to $u$ is $f^\prime(u)=e^{u}$.

Step6: Apply chain - rule

By the chain - rule $\frac{dy}{dx}=f^\prime(g(x))\cdot g^\prime(x)$. Substitute $u = g(x)=\sqrt{x}$ into $f^\prime(u)$ and multiply by $g^\prime(x)$. We get $\frac{dy}{dx}=e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}=\frac{e^{\sqrt{x}}}{2\sqrt{x}}$

Answer:

For $y = \sqrt[3]{1 + 7x}$, $(f(u),g(x))=(u^{\frac{1}{3}},1 + 7x)$ and $\frac{dy}{dx}=\frac{7}{3(1 + 7x)^{\frac{2}{3}}}$
For $y = e^{\sqrt{x}}$, $(f(u),g(x))=(e^{u},x^{\frac{1}{2}})$ and $\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{2\sqrt{x}}$