Answer to Question #149703 in Calculus for Maow

Question

Prove that if y= √x + 1/4 sin(2x^2)
dy/dx= (1/2√x) + 2xcos(2x)^2

Answer 1

Solution: Given that, $y=\sqrt{x}+\frac{1}{4} sin (2x^2)$

Now, we differentiate with respect to $x$

$\therefore \frac{dy}{dx}=\frac{d}{dx}[\sqrt{x}+\frac{1}{4} sin(2x^2)]$

$= \frac{d}{dx}[\sqrt{x}]+\frac{d}{dx}[\frac{1}{4} sin(2x^2)]$

$= \frac{d}{dx}[\sqrt{x}]+[\frac{1}{4} \frac{d}{dx}sin(2x^2)]$

$= \frac{d}{dx}[x^\frac{1}{2}]+[\frac{1}{4} \frac{d}{dx}sin(2x^2)]$

$= \frac{1}{2}x^{\frac{1}{2}-1}+[\frac{1}{4} cos(2x^2)\frac{d}{dx}(2x^2)] [Since \frac{d}{dx}x^n=nx^{n-1} and \frac{d}{dx} sin x= cos x ]$

$= \frac{1}{2}x^{-\frac{1}{2}}+[\frac{1}{4} cos(2x^2).2.\frac{d}{dx}(x^2)]$

$= \frac{1}{2x^{\frac{1}{2}}}+[\frac{1}{4}2 cos(2x^2).2.2x]$

$= \frac{1}{2\sqrt{x}}+[\frac{1}{4} 2cos(2x^2).4x]$

$= \frac{1}{2\sqrt{x}}+\frac{ 8xcos(2x^2)}{4}$ [Cancel out 4 with 4]

$\therefore\frac{dy}{dx}= \frac {1}{2\sqrt{x}}+2x cos(2x^2)$