Question

$\frac{d(\tan x.)}{dx}$

• A. ${\sec }^{2}x$
• B. ${\cot }^{2}x$
• C. ${\cos }^{2}x$
• D. ${\sin }^{2}x$

Answer 1

The correct option is A ${\sec }^{2}x$

$y=\tan x.$

$\therefore \frac{dy}{dx}=\underset{\delta x\to 0}{lim}\frac{\tan (x+\delta x)-\tan x}{\delta x}$
(Change to $\sin$ and $\cos$).

$=\underset{\delta x\to 0}{lim}\frac{\sin (x+\delta x-x)}{\delta x.\cos x\cos (x+\delta x)}$

$=\underset{\delta x\to 0}{lim}\frac{\sin \delta x}{\delta x}.\frac{1}{\cos x\cos (x+\delta x)}$

$=1.{\sec }^{2}x={\sec }^{2}x.$