Question

let $y = \frac{\sin(x)}{(8 \cdot x + 2)}$. find $\frac{dy}{dx}$. $\frac{dy}{dx}=$
type $\sin(x)$ for $\sin(x)$, $\cos(x)$ for $\cos(x)$, and so on.
type $(\sin(x))^2$ to square $\sin(x)$, $(f(x))^2$ to square $f(x)$.
use / to divide.
do not simplify your answer.

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u = \sin(x)$ and $v=8x + 2$.

Step2: Find $u'$ and $v'$

The derivative of $u=\sin(x)$ with respect to $x$ is $u'=\cos(x)$, and the derivative of $v = 8x+2$ with respect to $x$ is $v'=8$.

Step3: Substitute into quotient - rule formula

$\frac{dy}{dx}=\frac{\cos(x)(8x + 2)-\sin(x)\times8}{(8x + 2)^{2}}$

Answer:

$\frac{\cos(x)(8x + 2)-8\sin(x)}{(8x + 2)^{2}}$