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 $\frac{dy}{dx}$
$y = sin\frac{x}{3}$
identify the inner and outer functions. choose the correct answer
a. u = g(x) = sin x and y = f(u) = $\frac{u}{3}$
b. u = g(x) = sin x and y = f(u) = 3u
c. u = g(x) = $\frac{x}{3}$ and y = f(u) = sin u
d. u = g(x) = 3x and y = f(u) = sin u

Explanation:

Step1: Identify inner - outer functions

For the composite function $y = \sin\frac{x}{3}$, if we let $u = g(x)=\frac{x}{3}$ and $y = f(u)=\sin u$, we can write $y = f(g(x))$. So the correct answer for identifying inner and outer functions is C.

Step2: Apply chain - rule

The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$.
First, find $\frac{dy}{du}$:
Since $y = \sin u$, then $\frac{dy}{du}=\cos u$.
Second, find $\frac{du}{dx}$:
Since $u=\frac{x}{3}$, then $\frac{du}{dx}=\frac{1}{3}$.

Step3: Calculate $\frac{dy}{dx}$

Substitute $u = \frac{x}{3}$ back into $\frac{dy}{du}$ and multiply by $\frac{du}{dx}$:
$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\cos u\cdot\frac{1}{3}=\frac{1}{3}\cos\frac{x}{3}$.

Answer:

C. $u = g(x)=\frac{x}{3}$ and $y = f(u)=\sin u$, $\frac{dy}{dx}=\frac{1}{3}\cos\frac{x}{3}$