If y = sec(tan^{-1}x),then frac{dy}{dx} at x | vedclass

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

Question

(A) Given $y = \sec(	an^{-1}x)$.Let $	an^{-1}x = 	heta$,then $x = 	an	heta$.Since $\sec^2	heta = 1 + 	an^2	heta$,we have $\sec	heta = \sqrt{1

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) Given $y = \sec(	an^{-1}x)$.Let $	an^{-1}x = 	heta$,then $x = 	an	heta$.Since $\sec^2	heta = 1 + 	an^2	heta$,we have $\sec	heta = \sqrt{1 + 	an^2	heta} = \sqrt{1 + x^2}$.Thus,$y = \sqrt{1 + x^2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{1 + x^2}) = \frac{1}{2\sqrt{1 + x^2}} \cdot \frac{d}{dx}(1 + x^2) = \frac{1}{2\sqrt{1 + x^2}} \cdot 2x = \frac{x}{\sqrt{1 + x^2}}$.At $x = 1$:$\frac{dy}{dx} = \frac{1}{\sqrt{1 + (1)^2}} = \frac{1}{\sqrt{2}}$.

If $y = \sec(\tan^{-1}x)$,then $\frac{dy}{dx}$ at $x = 1$ is equal to:

Similar Questions

Assertion : Free electrons always keep on moving in a conductor even then no magnetic force acts on them in a magnetic field unless a current is passed through it.
Reason : The average velocity of free electrons is zero.

The upper half of an inclined plane of inclination $\theta$ is perfectly smooth while the lower half is rough. $A$ block starting from rest at the top of the plane will again come to rest at the bottom,if the coefficient of friction between the block and the lower half of the plane is given by:

The angle between the lines represented by $y^2 \sin^2 \theta - xy \sin^2 \theta + x^2(\cos^2 \theta - 1) = 0$ is

If the acute angle between the lines $ax^2+2hxy+by^2=0$ is $\frac{\pi}{4}$,then $4h^2=$

Which one of the following is the first step in taxonomy?

Vedclass Test Series

Exam Paper Generator

Online Exam Module