Question

find the derivative of $y = \sin(9x)\cdot\sin(7x)$
$\frac{dy}{dx}=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Here, let $u = \sin(9x)$ and $v=\sin(7x)$.

Step2: Find $\frac{du}{dx}$

Using the chain - rule, if $u=\sin(9x)$, let $t = 9x$, then $\frac{du}{dt}=\cos(t)$ and $\frac{dt}{dx}=9$. So, $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}=9\cos(9x)$.

Step3: Find $\frac{dv}{dx}$

Using the chain - rule, if $v = \sin(7x)$, let $s = 7x$, then $\frac{dv}{ds}=\cos(s)$ and $\frac{ds}{dx}=7$. So, $\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}=7\cos(7x)$.

Step4: Substitute into product - rule

$\frac{dy}{dx}=\sin(9x)\cdot7\cos(7x)+\sin(7x)\cdot9\cos(9x)=7\sin(9x)\cos(7x)+9\sin(7x)\cos(9x)$.

Answer:

$7\sin(9x)\cos(7x)+9\sin(7x)\cos(9x)$