Question

Differentiate the following w.r.t. x : ${\displaystyle {\cot }^{-1}\left(\frac{a^2-6x^2}{5ax}\right)}$

Answer 1

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

= ${\displaystyle {\tan }^{-1}\left(\frac{5ax}{a^2-6x^2}\right) ...[\because {\cot }^{-1}x={\tan }^{-1}\left(\frac{1}{x}\right)]}$

= ${\displaystyle {\tan }^{-1}\left[\frac{5\left(\frac{x}{a}\right)}{1-6{\left(\frac{x}{a}\right)}^2}\right]}$      ...[Dividing by a²]

= ${\displaystyle \tan \left[\frac{3\left(\frac{x}{a}\right)+2\left(\frac{x}{a}\right)}{1-3\left(\frac{x}{a}\right)\times 2\left(\frac{x}{a}\right)}\right]}$

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

Differentiating w.r.t. x, we get

${\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}[{\tan }^{-1}\left(\frac{3x}{a}\right)+{\tan }^{-1}\left(\frac{2x}{a}\right)]}$

= ${\displaystyle \frac{\text{d}}{\text{dx}}\left[{\tan }^{-1}\left(\frac{3x}{a}\right)\right]+\frac{\text{d}}{\text{dx}}\left[{\tan }^{-1}\left(\frac{2x}{a}\right)\right]}$

= ${\displaystyle \frac{1}{1+{\left(\frac{3x}{a}\right)}^2}\cdot \frac{d}{dx}\left(\frac{3x}{a}\right)+\frac{1}{1+{\left(\frac{2x}{a}\right)}^2}\cdot \frac{d}{dx}\left(\frac{2x}{a}\right)}$

= ${\displaystyle \frac{1}{1+\left(\frac{9x^2}{a^2}\right)}\times \frac{3}{a}\times 1+\frac{1}{1+\left(\frac{4x^2}{a^2}\right)}\times \frac{2}{a}\times 1}$

= ${\displaystyle \frac{a^2}{a^2+9x^2}\times \frac{3}{a}+\frac{a^2}{a^2+4x^2}\times \frac{2}{a}}$

= ${\displaystyle \frac{3a}{a^2+9x^2}+\frac{2a}{a^2+4x^2}}$