The differential coefficient of sec^{-1} x is | vedclass

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

Question

(C) The derivative of the inverse trigonometric function $\sec^{-1} x$ with respect to $x$ is a standard formula in calculus.By definition,if $y = \se

Vedclass · Accepted Answer

$\frac{1}{x\sqrt{x^2-1}}$

Vedclass · Answer

(C) The derivative of the inverse trigonometric function $\sec^{-1} x$ with respect to $x$ is a standard formula in calculus.By definition,if $y = \sec^{-1} x$,then $x = \sec y$.Differentiating both sides with respect to $y$,we get $\frac{dx}{dy} = \sec y 	an y$.Since $	an y = \sqrt{\sec^2 y - 1} = \sqrt{x^2 - 1}$,we have $\frac{dx}{dy} = x\sqrt{x^2 - 1}$.Therefore,$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{x\sqrt{x^2 - 1}}$.Thus,the correct option is $C$.

The differential coefficient of $\sec^{-1} x$ is:

Similar Questions

The differential coefficient of $\frac{{\tan^{-1}x}}{{1 + \tan^{-1}x}}$ with respect to $\tan^{-1}x$ is

Find the derivative of the following function (it is to be understood that $p, q, r$ are fixed non-zero constants): $\frac{p x^{2}+q x+r}{x}$

If $y=\sqrt{2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)}$,then find $\frac{d y}{d x}$ at $x=\frac{\pi}{4}$.

If $f(x) = \cos^{-1}\left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} \right]$,then the value of $f'(e)$ is:

Let $f(x)$ and $g(x)$ be differentiable functions on $R$. If $h(x) = f(g(f(x)))$,where $f(2) = 1$,$g(1) = 2$ and $f'(2) = g'(1) = 4$,then $h'(2)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module