Question

d/dx tan⁻¹x/1+tan⁻¹x w.r.t. tan⁻¹x=........,Select Proper option from the given options.
(a) 1/1+tan⁻¹x
(b) 1/(1+tan⁻¹x)²
(c) 1/1+x²
(d) -1/1+x²

Answer 1

we have to find the value of $\frac{d}{dx}\frac{tan^{-1}x}{1+tan^{-1}x}$ w.r.t to tan^{-1}x

first differentiate, $y=\frac{tan^{-1}x}{1+tan^{-1}x}$ w.r.t x ,


dy/dx = $\frac{(1+tan^{-1}x)\frac{d(tan^{-1}x)}{dx}-tan^{-1}x\frac{d(1+tan^{-1}x)}{dx}}{(1+tan^{-1}x)^2}$

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

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

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


now, Let z = tan^{-1}x

differentiate it with respect to x,

dz/dx = 1/(1 + x²)

now, we have to find dy/dz
dy/dz = dy/dx × dx/dz

= {dy/dx}/{dz/dx}

= 1/(1 + tan^-1x)²

hence, option (b) is correct.

Answer 2

Hello,

Solution:

let y = tan⁻¹x/1+tan⁻¹x

$y = \frac{ {tan}^{ - 1} x}{ 1+ {tan}^{ - 1} x}$
add and subtract 1 in numerator

$y = \frac{ {tan}^{ - 1}x + 1 - 1 }{1 + {tan}^{ - 1} x} \\ \\ y = \frac{1 + {tan}^{ - 1}x }{1 + {tan}^{ - 1} x} - \frac{1}{ 1+ {tan}^{ - 1} x} \\ \\y = 1 - \frac{1}{ 1+ {tan}^{ - 1} x} \\ \frac{dy}{dx} = 0 - \frac{(1 + {tan}^{ - 1}x) \frac{d(1)}{dx} - 1 \frac{d(1 + {tan}^{ - 1}x) }{dx} }{ {(1 + {tan}^{ - 1}x) }^{2} } \\ \frac{dy}{dx} = \frac{1}{(1 + {x}^{2}) ( {1 + {tan}^{ - 1}x) }^{2} }$
let m = tan⁻¹x

$\frac{dm}{dx} = \frac{1}{1 + {x}^{2} }$
Now to find

$\frac{dy}{dm} = \frac{ \frac{dy}{dx} }{ \frac{dm}{dx} } = \frac{1}{(1 + {x}^{2} )( {1 + {tan}^{ - 1}x) }^{2} } \times (1 + {x}^{2} ) \\ \\ \frac{dy}{dm} = \frac{1}{ {(1 + {tan}^{ - 1} x})^{2} }$
Option b is correct.

Hope it helps you.