Question

given the function $y = sqrt4{16x}cos x$, find $\frac{dy}{dx}$ in any form.

Explanation:

Step1: Rewrite the function

First, rewrite $\sqrt[4]{16x}$ as $(16x)^{\frac{1}{4}} = 2x^{\frac{1}{4}}$. So $y = 2x^{\frac{1}{4}}\cos x$.

Step2: Apply the product - rule

The product - rule states that if $y = u\cdot v$, where $u = 2x^{\frac{1}{4}}$ and $v=\cos x$, then $\frac{dy}{dx}=u'\cdot v + u\cdot v'$.
Find $u'$: Using the power - rule $\frac{d}{dx}(ax^{n})=anx^{n - 1}$, for $u = 2x^{\frac{1}{4}}$, $u'=2\times\frac{1}{4}x^{\frac{1}{4}-1}=\frac{1}{2}x^{-\frac{3}{4}}$.
Find $v'$: The derivative of $\cos x$ is $-\sin x$.

Step3: Substitute $u$, $u'$, $v$, and $v'$ into the product - rule

$\frac{dy}{dx}=\frac{1}{2}x^{-\frac{3}{4}}\cos x+2x^{\frac{1}{4}}(-\sin x)$.
Simplify to get $\frac{dy}{dx}=\frac{\cos x}{2x^{\frac{3}{4}}}-2x^{\frac{1}{4}}\sin x$.

Answer:

$\frac{\cos x}{2x^{\frac{3}{4}}}-2x^{\frac{1}{4}}\sin x$