If x = e^{tan^{-1}left(frac{y-x^2}{x^2}right) | vedclass

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

Question

(D) Given the equation $x = e^{	an^{-1}\left(\frac{y-x^2}{x^2}ight)}$.Taking the natural logarithm on both sides,we get $\ln(x) = 	an^{-1}\left(\f

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Given the equation $x = e^{	an^{-1}\left(\frac{y-x^2}{x^2}ight)}$.Taking the natural logarithm on both sides,we get $\ln(x) = 	an^{-1}\left(\frac{y-x^2}{x^2}ight)$.This implies $	an(\ln x) = \frac{y-x^2}{x^2}$.Rearranging the terms,we have $y = x^2 	an(\ln x) + x^2$.Differentiating both sides with respect to $x$ using the product rule and chain rule:$\frac{dy}{dx} = \frac{d}{dx}[x^2 	an(\ln x)] + \frac{d}{dx}[x^2]$$\frac{dy}{dx} = [2x 	an(\ln x) + x^2 \cdot \sec^2(\ln x) \cdot \frac{1}{x}] + 2x$$\frac{dy}{dx} = 2x 	an(\ln x) + x \sec^2(\ln x) + 2x$.Now,evaluate at $x = 1$:$\frac{dy}{dx} \Big|_{x=1} = 2(1) 	an(\ln 1) + 1 \sec^2(\ln 1) + 2(1)$Since $\ln 1 = 0$ and $	an 0 = 0$ and $\sec 0 = 1$:$\frac{dy}{dx} \Big|_{x=1} = 2(0) + 1(1)^2 + 2 = 0 + 1 + 2 = 3$.

If $x = e^{\tan^{-1}\left(\frac{y-x^2}{x^2}\right)}$,then $\frac{dy}{dx}$ at $x = 1$ is

Similar Questions

If $e^{-y} \cdot y = x$,then $\frac{dy}{dx}$ is

If $\log (x+y)=\sin (x+y)$,then $\frac{dy}{dx}$ is

If $\sin y = x \cos (a + y),$ then $\frac{dy}{dx} = $

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

If $xy \neq 0, x+y \neq 0$ and $x^m y^n=(x+y)^{m+n}$,where $m, n \notin N$,then $\frac{dy}{dx}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module