Question

given y = f(u) and u = g(x), find $\frac{dy}{dx}=f(g(x))g(x)$ for the following functions.
y = sin u, u = 3x + 2
$\frac{dy}{dx}=f(g(x))g(x)=square$

Explanation:

Step1: Find the derivative of y with respect to u

The derivative of $y = \sin u$ with respect to $u$ is $y^\prime=f^\prime(u)=\cos u$.

Step2: Find the derivative of u with respect to x

The derivative of $u = 3x + 2$ with respect to $x$ is $u^\prime=g^\prime(x)=3$.

Step3: Apply the chain - rule

By the chain - rule $\frac{dy}{dx}=f^\prime(g(x))g^\prime(x)$. Substitute $u = g(x)=3x + 2$ into $f^\prime(u)$ and multiply by $g^\prime(x)$. We get $\frac{dy}{dx}=\cos(3x + 2)\times3 = 3\cos(3x + 2)$.

Answer:

$3\cos(3x + 2)$