Question

find $\frac{dy}{dx}$ for the following function. $y = \frac{9sin x}{9 - 4sin x}$

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, $u = 9\sin x$ and $v=9 - 4\sin x$. First, find $u'$ and $v'$. The derivative of $u = 9\sin x$ with respect to $x$ is $u'=9\cos x$, and the derivative of $v = 9-4\sin x$ with respect to $x$ is $v'=- 4\cos x$.

Step2: Substitute into quotient - rule formula

$\frac{dy}{dx}=\frac{(9\cos x)(9 - 4\sin x)-(9\sin x)(-4\cos x)}{(9 - 4\sin x)^{2}}$.

Step3: Expand the numerator

Expand the numerator: $(9\cos x)(9 - 4\sin x)-(9\sin x)(-4\cos x)=81\cos x-36\sin x\cos x + 36\sin x\cos x$.

Step4: Simplify the numerator and get the result

The $-36\sin x\cos x$ and $36\sin x\cos x$ in the numerator cancel out. So, $\frac{dy}{dx}=\frac{81\cos x}{(9 - 4\sin x)^{2}}$.

Answer:

$\frac{81\cos x}{(9 - 4\sin x)^{2}}$