Question

If $y={\tan }^{-1}\left(\frac{5ax}{a^2-6x^2}\right)$, prove that:
$\frac{dy}{dx}=\frac{3a}{a^2+9x^2}+\frac{2a}{a^2+4x^2}$.

Answer 1

Given

$y={\tan }^{-1}\left(\frac{5ax}{a^2-6x^2}\right)$

or, $y={\tan }^{-1}\left(\frac{\frac{5x}{a}}{1-\frac{6x^2}{a^2}}\right)$ [ Dividing the numerator and the denominator by $a^2$ we get,]

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

or, $y={\tan }^{-1}\frac{3x}{a}+{\tan }^{-1}\frac{2x}{a}$

Now differentiating both sides with respect to $x$ we get,

$\frac{dy}{dx}=\frac{a^2}{a^2+9x^2}.\frac{3}{a}+\frac{a^2}{a^2+4x^2}.\frac{2}{a}$

or, $\frac{dy}{dx}=\frac{3a}{a^2+9x^2}+\frac{2a}{a^2+4x^2}$