Discussion Forum

Derivative of log (sec $\theta$+ tan $\theta$)  with respect to $\sec \theta$  at $\theta$ =$\frac{\pi}{4}$

Answer:

Explanation:

Let u=log (sec $\theta$+ tan $\theta$)   and v= $\sec \theta$

 On differentiating both sides w.r.t  $\theta$   , we get

$\frac{du}{d\theta}=\frac{1}{(\sec \theta+\tan \theta)}(\sec \theta\tan \theta+\sec^{2} \theta)$  and 

$\frac{dv}{d\theta}=\sec \theta\tan \theta$

 $\therefore$  $\frac{dv}{d\theta}=\frac{\frac{du}{d\theta}}{\frac{dv}{d\theta}}$

 =  $\frac{\sec \theta(\tan \theta+\sec \theta)}{( \sec\theta+\tan \theta)\times \sec \theta \tan \theta}= \cot \theta$

$\Rightarrow$    $\frac{du}{dv(\theta=\pi/4)}=\cot \frac{\pi}{4}=1$