Question

find $\frac{dy}{dx}$ for the following function.
y = $\frac{8cos x}{6 - 7cos x}$
$\frac{dy}{dx}=square$

Explanation:

Step1: Identify quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u = 8\cos x$ and $v=6 - 7\cos x$.

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

Differentiate $u = 8\cos x$ with respect to $x$. Using the derivative formula $\frac{d}{dx}(\cos x)=-\sin x$, we get $u'=-8\sin x$.
Differentiate $v = 6-7\cos x$ with respect to $x$. The derivative of a constant is 0 and $\frac{d}{dx}(- 7\cos x)=7\sin x$, so $v' = 7\sin x$.

Step3: Apply the quotient - rule

Substitute $u$, $u'$, $v$, and $v'$ into the quotient - rule formula:
\[

$$\begin{align*} \frac{dy}{dx}&=\frac{(-8\sin x)(6 - 7\cos x)-8\cos x(7\sin x)}{(6 - 7\cos x)^{2}}\\ &=\frac{-48\sin x + 56\sin x\cos x-56\sin x\cos x}{(6 - 7\cos x)^{2}}\\ &=\frac{-48\sin x}{(6 - 7\cos x)^{2}} \end{align*}$$

\]

Answer:

$\frac{-48\sin x}{(6 - 7\cos x)^{2}}$