Question

differentiate.
y = \frac{4x}{5 - \tan(x)}
y=

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1. - / 1 points

differentiate.
f(\theta)=\theta\cos(\theta)\sin(\theta)
f(\theta)=

Explanation:

Step1: Apply quotient - rule for $y=\frac{4x}{5 - \tan(x)}$

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 4x$, so $u'=4$; $v = 5-\tan(x)$, so $v'=-\sec^{2}(x)$. Then $y'=\frac{4(5 - \tan(x))-4x(-\sec^{2}(x))}{(5 - \tan(x))^{2}}=\frac{20 - 4\tan(x)+4x\sec^{2}(x)}{(5 - \tan(x))^{2}}$.

Step2: Apply product - rule for $f(\theta)=\theta\cos(\theta)\sin(\theta)$

First, rewrite $f(\theta)$ as $f(\theta)=\frac{1}{2}\theta\sin(2\theta)$ (using the double - angle formula $\sin(2\theta)=2\sin(\theta)\cos(\theta)$).
The product - rule states that if $y = uv$, then $y'=u'v+uv'$. Let $u=\frac{1}{2}\theta$, so $u'=\frac{1}{2}$; $v = \sin(2\theta)$, so $v'=2\cos(2\theta)$.
Then $f'(\theta)=\frac{1}{2}\sin(2\theta)+\frac{1}{2}\theta\times2\cos(2\theta)=\frac{1}{2}\sin(2\theta)+\theta\cos(2\theta)$.

Answer:

$y'=\frac{20 - 4\tan(x)+4x\sec^{2}(x)}{(5 - \tan(x))^{2}}$
$f'(\theta)=\frac{1}{2}\sin(2\theta)+\theta\cos(2\theta)$