If y = (1 - x^2) operatorname{Tanh}^{-1} x,th | vedclass

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

Question

(D) Given $y = (1 - x^2) \operatorname{Tanh}^{-1} x$. First,differentiate with respect to $x$: $\frac{dy}{dx} = (1 - x^2) \cdot \frac{d}{dx}(\operator

Vedclass · Accepted Answer

$-\frac{2(x+y)}{1-x^2}$

Vedclass · Answer

(D) Given $y = (1 - x^2) \operatorname{Tanh}^{-1} x$. First,differentiate with respect to $x$: $\frac{dy}{dx} = (1 - x^2) \cdot \frac{d}{dx}(\operatorname{Tanh}^{-1} x) + \operatorname{Tanh}^{-1} x \cdot \frac{d}{dx}(1 - x^2)$ $\frac{dy}{dx} = (1 - x^2) \cdot \frac{1}{1 - x^2} + \operatorname{Tanh}^{-1} x \cdot (-2x)$ $\frac{dy}{dx} = 1 - 2x \operatorname{Tanh}^{-1} x$. Now,differentiate again with respect to $x$: $\frac{d^2 y}{dx^2} = \frac{d}{dx}(1) - 2 \cdot \frac{d}{dx}(x \operatorname{Tanh}^{-1} x)$ $\frac{d^2 y}{dx^2} = 0 - 2 \left[ x \cdot \frac{1}{1 - x^2} + \operatorname{Tanh}^{-1} x \cdot 1 \right]$ $\frac{d^2 y}{dx^2} = -2 \left[ \frac{x}{1 - x^2} + \operatorname{Tanh}^{-1} x \right]$. From the original equation,$\operatorname{Tanh}^{-1} x = \frac{y}{1 - x^2}$. Substituting this into the expression: $\frac{d^2 y}{dx^2} = -2 \left[ \frac{x}{1 - x^2} + \frac{y}{1 - x^2} \right]$ $\frac{d^2 y}{dx^2} = -\frac{2(x + y)}{1 - x^2}$.

If $y = (1 - x^2) \operatorname{Tanh}^{-1} x$,then $\frac{d^2 y}{d x^2} = $

Similar Questions

If $y = a{x^{n + 1}} + b{x^{ - n}},$ then ${x^2}\frac{{{d^2}y}}{{d{x^2}}}$ is equal to:

The $n^{th}$ derivative of $x e^x$ vanishes when

Let $f(x)$ and $g(x)$ be two functions having finite non-zero $3^{rd}$ order derivatives $f'''(x)$ and $g'''(x)$ for all $x \in R$. If $f(x)g(x) = 1$ for all $x \in R$,then $\frac{f'''}{f'} - \frac{g'''}{g'}$ is equal to

If $y = \frac{1}{x} + \cos 2x$,then $\frac{d^2y}{dx^2}$ is equal to

If $y = \frac{A}{x} + B x^2$,then $x^2 \frac{d^2 y}{d x^2} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module