Question

If y=tan^-1 1+x²/1-x² then find dy/dx=? ​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{y=tan^{-1}\left(\dfrac{1+x^2}{1-x^2}\right)}$

$\underline{\textbf{To find:}}$

$\mathsf{\dfrac{dy}{dx}}$

$\underline{\textbf{Solution:}}$

$\textsf{We apply chain rule to find the derivative}$

$\mathsf{Consider,}$

$\mathsf{y=tan^{-1}\left(\dfrac{1+x^2}{1-x^2}\right)}$

$\textsf{Differentiate with respect to x}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{1+\left(\dfrac{1+x^2}{1-x^2}\right)^2}\;\dfrac{d\left(\dfrac{1+x^2}{1-x^2}\right)}{dx}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(1-x^2)^2}{(1-x^2)^2+(1+x^2)^2}\;\left(\dfrac{(1-x^2)2x-(1+x^2)(-2x)}{(1-x^2)^2}\right)}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(1-x^2)^2}{(1-x^2)^2+(1+x^2)^2}\;\left(\dfrac{(1-x^2)2x+(1+x^2)(2x)}{(1-x^2)^2}\right)}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{(1-x^2)^2+(1+x^2)^2}\;\left(\dfrac{2x(1-x^2+1+x^2)}{1}\right)}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{(1-x^2)^2+(1+x^2)^2}\;(2x(2))}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{1+x^4-2x^2+1+x^4+2x^2}\;(4x)}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{4x}{2+2x^4}}$

$\implies\boxed{\mathsf{\dfrac{dy}{dx}=\dfrac{2x}{1+x^4}}}$

$\underline{\textbf{Find more:}}$

If y = tan-1(6x-7/6+7x)

then dy/dx is

​https://brainly.in/question/36894600  

Answer 2

Given :  

           $\tan x^-11 +\frac{x^{2} }{1-x^2}$

Find :

            $\frac{dy}{dx}$ = $\frac{\partial \left ( tanx^-11+\frac{x^2}{1-x^2} \right )}{\partial x}$

Step-by-step explanation:

$\frac{dy}{dx} =$  $\frac{\partial \left ( tanx^-11+\frac{x^2}{1-x^2} \right )}{\partial x}$

By the rule of addition of derivative,

$\frac{dy}{dx} =$ $\frac{\partial \left ( tanx^-11 \right )}{\partial x}+\frac{\partial \left ( \frac{x^2}{1-x^2} \right )}{\partial x}$

$\frac{dy}{dx} =$ $nx^(-11-1)sec^2x^(-11) + \frac{[(1-x^2)*2x]-[(x^2)*(-2x)]}{(1-x^2)^2}$

∵    tanx^n dx = nx^(n-1)sec^2x^n &

     by the rule of division of deritative,

      $\frac{\partial [\frac{y}{x}]}{\partial x} = \frac{y*\frac{dx}{dy} - x*\frac{dy}{dx} }{ x^2}$

∴by solving above equation we get,

$\frac{dy}{dx} =$ $nx^(-12)sec^2x^2(-11) + \frac{2x-2x^3+2x^3}{(1+x^2)^2}$

$\frac{dy}{dx} =$  $nx^(-12)sec^2x^(-11) + \frac{2}{(1+x^2)^2}$

Final ans:

$\frac{dy}{dx} =$ $nx^(-12)sec^2x^(-11) + \frac{2}{(1+x^2)^2}$