Question

Prove that the $\frac{d}{dx}\left(secx\right)=secx\tan x$

Answer 1

STEP 1 : Solving the LHS of the equation

We know that,

$secx=\frac{1}{\cos x}$

Taking the LHS of the equation and applying quotient rule to find the derivative of

$\Rightarrow \frac{d}{dx}\left(secx\right)=\frac{d}{dx}\left(\frac{1}{\cos x}\right)$

$\Rightarrow \frac{{\displaystyle \frac{d}{dx}}1.\left(\cos x\right)-{\displaystyle \frac{d}{dx}}\cos x.\left(1\right)}{{\left(\cos x\right)}^2}$

Simplifying the above equation

$\Rightarrow \frac{\left(0\right)\left(\cos x\right)-\left(-\sin x\right)}{{\left(\cos x\right)}^2}$

$\Rightarrow \frac{\sin x}{\left(\cos x\right)\left(\cos x\right)}$

$\Rightarrow \left(\frac{\sin x}{\cos x}\right)\left(\frac{1}{\cos x}\right)$

$\Rightarrow \tan xsecx$

Thus, LHS = RHS.

Hence proved.