Question

Given $f\left(x\right)=\tan x$. Then $f^{\prime }\left(x\right)$ is equal to

• A. ${\sec }^{2}x$
• B. ${\text{cosec}}^{2}x$
• C. $-{\sec }^{2}x$
• D. $x{\sec }^{2}x$

Answer 1

The correct option is A

${\sec }^{2}x$

$f\left(x\right)=\tan x$

We can write
$\tan x$ as $\frac{\sin x}{\cos x}$

$\therefore f\left(x\right)=\frac{\sin x}{\cos x}=\frac{g\left(x\right)}{h\left(x\right)}$

We know,
$f^{\prime }\left(x\right)=\frac{h\left(x\right)g^{\prime }\left(x\right)-h^{\prime }\left(x\right)g\left(x\right)}{{\left(h\left(x\right)\right)}^2}$
$=\frac{\cos x{\sin }^{\prime }\left(x\right)-{\cos }^{\prime }\left(x\right)\sin x}{{\left(\cos x\right)}^2}$

$=-\frac{\cos x\cos x-(-\sin x)\left(\sin x\right)}{(\cos x{)}^2}$
$=\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}$
$={\sec }^2$
$\therefore f^{\prime }\left(x\right)={\sec }^{2}x\forall x\in (\frac{-\pi }{2},\frac{\pi }{2})$