Answer to Question #114646 in Calculus for ANJU JAYACHANDRAN

Question

Find the derivative of In(1+x^2) w.r.t tan inverse x

Answer 1

Let us assume that:

$y = \ln (1 + x^2)$

And:

$v = \tan^{-1} (x)$

Derivative of $y$ with respect to $x$ is:

$\frac{dy}{dx} = \frac{d}{dx} \left( \ln (1 + x^2) \right) \\ \hspace{0.5 cm} = \frac{1}{1+x^2} \frac{d}{dx} (1 + x^2) \\$

$\hspace{0.5 cm} = \frac{1}{1 + x^2} \left[ \frac{d}{dx} (1) + \frac{d}{dx} (x^2) \right]$

$\hspace{0.5 cm} = \frac{1}{1+x^2} \left( 0 + 2x \right) \hspace{1 cm}$ [ $\because \frac{d}{dx} (x^n) = nx^{n-1} \, and \, \frac{d}{dx} (c) = 0 \,$ ( Where c is a constant ) ]

$\hspace{0.5 cm} = \frac{2x}{1 + x^2}$

Derivative of $v$ with respect to $x$ is:

$\frac{dv}{dx} = \frac{d}{dx} \left( \tan^{-1} (x) \right) \\ \hspace{0.5 cm} = \frac{1}{1+x^2}$ [ $\because \frac{d}{dx} \left( \tan^{-1} (x) \right) = \frac{1}{1+x^2}$ ]

Now:

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

Hence:

The derivative of $\ln (1 + x^2)$ with respect to $\tan^{-1} (x)$ or derivative of $y$ with respect to $v$ is:

$\frac{dy}{dv} = 2x$