If y=e^{tan ^{-1} x}, then: | vedclass

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

Question

(A) Given $y=e^{	an ^{-1} x}$.Differentiating with respect to $x$, we get:$\frac{dy}{dx} = e^{	an ^{-1} x} 	imes \frac{d}{dx}(	an ^{-1} x) = e^{

Vedclass · Accepted Answer

$(1+x^2) y_2+(2 x-1) y_1=0$

Vedclass · Answer

(A) Given $y=e^{	an ^{-1} x}$.Differentiating with respect to $x$, we get:$\frac{dy}{dx} = e^{	an ^{-1} x} 	imes \frac{d}{dx}(	an ^{-1} x) = e^{	an ^{-1} x} 	imes \frac{1}{1+x^2}$.Since $y = e^{	an ^{-1} x}$, we can write:$\frac{dy}{dx} = \frac{y}{1+x^2}$.Multiplying both sides by $(1+x^2)$, we get:$(1+x^2) y_1 = y$.Differentiating again with respect to $x$ using the product rule on the left side:$\frac{d}{dx}[(1+x^2) y_1] = \frac{d}{dx}(y)$.$(1+x^2) y_2 + y_1(2x) = y_1$.Rearranging the terms:$(1+x^2) y_2 + (2x - 1) y_1 = 0$.

If $y=e^{\tan ^{-1} x}$, then:

Similar Questions

Let $f(x) = \tan^{-1} x$. Then $f'(x) + f''(x) = 0$ when $x$ is equal to:

If $y = a{e^{mx}} + b{e^{ - mx}}$,then $\frac{d^2y}{dx^2} - m^2y = $

If $f: R \rightarrow R$ is an even function which is twice differentiable on $R$ and $f^{\prime \prime}(\pi)=1$, then $f^{\prime \prime}(-\pi)$ is equal to

If $y = e^{nx}$,then $\left( \frac{d^2y}{dx^2} \right) \left( \frac{d^2x}{dy^2} \right)$ is equal to

If $f(x) = \sin x + \cos x$, then $f\left(\frac{\pi}{4}\right) f^{(iv)}\left(\frac{\pi}{4}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module