Question

Plzz ans differentiation​

Answer 1

Differential coefficient means derivative (I googled that up)

We are given the function

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

we have to differentiate it wrt x².

Let us take

$U = tan^{-1} \frac{2x}{1-x^2}$

$Let,x = tan\theta$

$So,U = tan^{-1} \frac{2tan\theta}{1-tan^2\theta}$

$U = tan^{-1}(tan2 \theta )$

$U = 2 \theta$

$U = 2tan^{-1}x$

$\frac{dU}{dx} = \frac{2}{1+x^2}$

Let us take

$V = x^2$

$\frac{dV}{dx} = 2x$

What we actually need to find is $\frac{dU}{dV}$

$\frac{dU}{dV} = \frac{dU}{dx} \times \frac{dx}{dV}\\ \\ \frac{dU}{dV} = \frac{2}{1+x^2}\times\frac{1}{2x} \\\\\large\black\boxed{\frac{dU}{dV} = \frac{1}{x+x^3}}$