Question

Find the derivative of $secx$.

Answer 1

Compute the derivative:

We know that

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

So we can find the derivative as

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

We know the quotient rule, $f\text{'}\left(\frac{u}{v}\right)=\frac{u\text{'}v-uv\text{'}}{v^2}$. That is here

$u=1$ and $v=\cos x$

$\Rightarrow u\text{'}=0$ and $v\text{'}=-\sin x$

So, the derivative can be found as

$f\text{'}\left(\frac{1}{\cos x}\right)=\frac{0\times \cos x-1\times \left(-\sin x\right)}{{\left(\cos x\right)}^2}$

$$\begin{gathered} =\frac{0+\sin x}{{\cos }^{2}x} \\ =\frac{\sin x}{\cos x}\times \frac{1}{\cos x} \\ =\tan x·secx \end{gathered}$$

Hence, the required derivative is $\tan x·secx$.