Question

for the following composite function, find an inner function u = g(x) and an outer function y = f(u) such that y = f(g(x)). then calculate $\frac{dy}{dx}$. y = sin $\frac{x}{18}$. identify the inner and outer functions. choose the correct answer. a. u = g(x) = 18x and y = f(u) = sin u b. u = g(x) = sin x and y = f(u) = $\frac{u}{18}$ c. u = g(x) = sin x and y = f(u) = 18u d. u = g(x) = $\frac{x}{18}$ and y = f(u) = sin u $\frac{dy}{dx}=square$

Explanation:

Step1: Identify inner and outer functions

For $y = \sin\frac{x}{18}$, let $u = g(x)=\frac{x}{18}$ and $y = f(u)=\sin u$. So the correct option is D.

Step2: Differentiate outer - function with respect to u

The derivative of $y = f(u)=\sin u$ with respect to $u$ is $\frac{dy}{du}=\cos u$.

Step3: Differentiate inner - function with respect to x

The derivative of $u = g(x)=\frac{x}{18}$ with respect to $x$ is $\frac{du}{dx}=\frac{1}{18}$.

Step4: Use chain - rule

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=\cos u$ and $\frac{du}{dx}=\frac{1}{18}$ into the chain - rule formula. Since $u = \frac{x}{18}$, we have $\frac{dy}{dx}=\cos\frac{x}{18}\cdot\frac{1}{18}=\frac{1}{18}\cos\frac{x}{18}$.

Answer:

D. $u = g(x)=\frac{x}{18}$ and $y = f(u)=\sin u$
$\frac{dy}{dx}=\frac{1}{18}\cos\frac{x}{18}$