Question

Differentiate the following w.r.t. x : ${\displaystyle {\tan }^{-1}\left(\frac{\cos 7x}{1+\sin 7x}\right)}$

Answer 1

Let y = ${\displaystyle {\tan }^{-1}\left(\frac{\cos 7x}{1+\sin 7x}\right)}$

= ${\displaystyle {\tan }^{-1}\left[\frac{\sin (\frac{\pi }{2}-7x)}{1+\cos (\frac{\pi }{2}-7x)}\right]}$

= ${\displaystyle {\tan }^{-1}\left[\frac{2\sin (\frac{\pi }{4}-\frac{7x}{2}).\cos (\frac{\pi }{4}-\frac{7x}{2})}{2{\cos }^2(\frac{\pi }{4}-\frac{7x}{2})}\right]}$

= ${\displaystyle {\tan }^{-1}\left[\tan (\frac{\pi }{4}-\frac{7x}{2})\right]}$

= ${\displaystyle \frac{\pi }{4}-\frac{7x}{2}}$
Differentiating w.r.t. x, we get
${\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\frac{\pi }{4}-\frac{7x}{2})}$

= ${\displaystyle \frac{\text{d}}{\text{dx}}\left(\frac{\pi }{4}\right)-\frac{7}{2}\frac{\text{d}}{\text{dx}}\left(x\right)}$

= ${\displaystyle 0-\frac{7}{2}\times 1}$

= ${\displaystyle -\frac{7}{2}}$.