If y = (1 + x^2)tan^{-1}x - x, then frac{dy}{ | vedclass

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

Question

(B) Given the function $y = (1 + x^2)	an^{-1}x - x.$ To find $\frac{dy}{dx},$ we apply the product rule to the first term $(1 + x^2)	an^{-1}x$ and t

Vedclass · Accepted Answer

$2x	an^{-1}x$

Vedclass · Answer

(B) Given the function $y = (1 + x^2)	an^{-1}x - x.$ To find $\frac{dy}{dx},$ we apply the product rule to the first term $(1 + x^2)	an^{-1}x$ and the power rule to the second term $-x.$ Using the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx},$ where $u = (1 + x^2)$ and $v = 	an^{-1}x$: $\frac{dy}{dx} = \frac{d}{dx}[(1 + x^2)	an^{-1}x] - \frac{d}{dx}(x)$ $\frac{dy}{dx} = (1 + x^2) \cdot \frac{d}{dx}(	an^{-1}x) + 	an^{-1}x \cdot \frac{d}{dx}(1 + x^2) - 1$ Since $\frac{d}{dx}(	an^{-1}x) = \frac{1}{1 + x^2}$ and $\frac{d}{dx}(1 + x^2) = 2x$: $\frac{dy}{dx} = (1 + x^2) \cdot \frac{1}{1 + x^2} + 	an^{-1}x \cdot (2x) - 1$ $\frac{dy}{dx} = 1 + 2x	an^{-1}x - 1$ $\frac{dy}{dx} = 2x	an^{-1}x.$ Thus,the correct option is $B$.

If $y = (1 + x^2)\tan^{-1}x - x,$ then $\frac{dy}{dx} = $

Similar Questions

For some constants $a$ and $b$,find the derivative of $\frac{x-a}{x-b}$.

$\frac{d}{dx}(e^{x^3})$ is equal to

$\frac{d}{dx}\sqrt{\sec^2 x + \text{cosec}^2 x} = $

Evaluate: $\frac{d}{d x}\left[e^{\tan ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]$

If $f(1) = 1$ and $f'(1) = 3$,then the derivative of $f(f(f(x))) + (f(x))^2$ at $x = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module