If y = (tan^{-1} x)^2 ,then (x^2 + 1)^2 y_2 | vedclass

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

Question

(D) Given $ y = (	an^{-1} x)^2 $.First,differentiate with respect to $ x $:$ y_1 = \frac{dy}{dx} = 2(	an^{-1} x) \cdot \frac{1}{1+x^2} $.This implie

Vedclass · Accepted Answer

$ 2 $

Vedclass · Answer

(D) Given $ y = (	an^{-1} x)^2 $.First,differentiate with respect to $ x $:$ y_1 = \frac{dy}{dx} = 2(	an^{-1} x) \cdot \frac{1}{1+x^2} $.This implies $ (1+x^2) y_1 = 2 	an^{-1} x $.Differentiate both sides again with respect to $ x $ using the product rule:$ (1+x^2) y_2 + y_1(2x) = 2 \cdot \frac{1}{1+x^2} $.Multiply the entire equation by $ (1+x^2) $:$ (1+x^2)^2 y_2 + 2x(1+x^2) y_1 = 2 $.Thus,the value is $ 2 $.

If $ y = (\tan^{-1} x)^2 $,then $ (x^2 + 1)^2 y_2 + 2x(x^2 + 1) y_1 $ is equal to

Similar Questions

If $x=\sin \theta$ and $y=\sin(k \theta)$, then $(1-x^2) y_2-x y_1-\alpha y=0$, for $\alpha=$

$\frac{d^2x}{dy^2} = $

Let $f(x) = x^{3} + x^{2}f^{\prime}(1) + 2x f^{\prime\prime}(2) + f^{\prime\prime\prime}(3)$, where $x \in R$. Then the value of $f^{\prime}(5)$ is:

Find the second order derivative of the function $\log (\log x)$.

If $y=e^{ax}(\cos bx+\sin bx)$ satisfies the equation $\frac{d^2y}{dx^2}-K\frac{dy}{dx}+Ly=0$, then $L+bK=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module