frac{d}{dx}left( frac{1}{x^4 sec x} right) = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that $\frac{1}{\sec x} = \cos x$. Therefore,the expression becomes $\frac{d}{dx}\left( \frac{\cos x}{x^4} \right)$. Using the quotient rul

Vedclass · Accepted Answer

$\frac{-(x \sin x + 4 \cos x)}{x^5}$

Vedclass · Answer

(B) We know that $\frac{1}{\sec x} = \cos x$. Therefore,the expression becomes $\frac{d}{dx}\left( \frac{\cos x}{x^4} \right)$. Using the quotient rule $\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = \cos x$ and $v = x^4$: $\frac{d}{dx}\left( \frac{\cos x}{x^4} \right) = \frac{x^4(-\sin x) - \cos x(4x^3)}{(x^4)^2}$ $= \frac{-x^4 \sin x - 4x^3 \cos x}{x^8}$ $= \frac{-x^3(x \sin x + 4 \cos x)}{x^8}$ $= \frac{-(x \sin x + 4 \cos x)}{x^5}$.

$\frac{d}{dx}\left( \frac{1}{x^4 \sec x} \right) = $

Similar Questions

$\frac{d}{dx} \sqrt{\sin 2x}$ is equal to

If $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + .....\infty$,then $\frac{dy}{dx} = $

The value of $\frac{d}{dx}(\log_e x)$ is:

At point $P$ on the given curve,the value of $\frac{dy}{dx}$ is:

The side of a square is increasing at the rate of $0.2\,cm/s$. The rate of increase of perimeter with respect to time is $...........\,cm/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module