Question

Find the first derivative of the function square root of x+1/4sin(2x)squared

Answer 1

I'm thinking you want something that looks like this
$\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}$ or it could be like this
$\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)$

use chain rule
dy/dx f(g(x))=f'(g(x))g'(x)
and √x=x^(1/2)
so
I'll do the first one first
$\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}$=
$\frac{dy}{dx}(x+\frac{1}{4}sin^2(2x))^{\frac{1}{2}}$=
$\frac{1}{2(x+\frac{1}{4}sin^2(2x))^{\frac{1}{2}}}(1+sin(2x)cos(2x))$=
$\frac{1+sin(2x)cos(2x)}{2\sqrt{x+\frac{1}{4}sin^2(2x}}$

the 2nd one is a bit simpler
$\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)$=
$\frac{dy}{dx}x^{\frac{1}{2}}+\frac{1}{4}sin^2(2x)$=
$\frac{1}{2}x^{\frac{-1}{2}}+\frac{1}{4}sin^2(2x)$=
$\frac{1}{2x^{\frac{1}{2}}}+sin(2x)cos(2x)$=
$\frac{1}{2\sqrt{x}}+sin(2x)cos(2x)$


in conclusion
$\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}$=$\frac{1+sin(2x)cos(2x)}{2\sqrt{x+\frac{1}{4}sin^2(2x}}$ and
$\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)$=$\frac{1}{2\sqrt{x}}+sin(2x)cos(2x)$
depends which one it was