Question

If $y={\tan }^{-1}\left(\frac{3a^{2}x-x^3}{a^3-3a{x}^2}\right)$ then $\frac{dy}{dx}=$?

• A. $\frac{3a}{a^2+x^2}$
• B. $\frac{1}{a^2+x^2}$
• C. $\frac{-3a^2}{a^2+x^2}$
• D. $\frac{-3a}{a^2+x^2}$

Answer 1

The correct option is A $\frac{3a}{a^2+x^2}$

Consider the given expression.

$y={\tan }^{-1}\left(\frac{3a^{2}x-x^3}{a^3-3a{x}^2}\right)$

Let,

$x=a\tan \theta$

$\theta ={\tan }^{-1}\frac{x}{a}$

Therefore,

$y={\tan }^{-1}\left[\frac{a^3(3\tan \theta -{\tan }^3\theta )}{a^3(1-3{\tan }^2\theta )}\right]$

$y={\tan }^{-1}\left(\tan 3\theta \right)$

$y=3\theta$

$y=3{\tan }^{-1}\frac{x}{a}$

Differentiate $y$ with respect to $x$.

$\frac{dy}{dx}=3\times \frac{1}{1+{\left(\frac{x}{a}\right)}^2}\times \frac{1}{a}$

$\frac{dy}{dx}=\frac{3a}{a^2+x^2}$

Hence, this is the required result.