# How do you differentiate f(x)=sec(1/sqrtx )  using the chain rule?

Dec 7, 2016

$\left(- \frac{1}{2}\right) \left(\frac{1}{x} ^ \left(\frac{3}{2}\right)\right) \sec \left(\frac{1}{\sqrt{x}}\right) \tan \left(\frac{1}{\sqrt{x}}\right)$

#### Explanation:

let $\frac{1}{\sqrt{x}}$ be 'p' which is obviously be a function of x. Thus sec (p) is to be differentiated w.r.t to x, that is to find out $\frac{d}{\mathrm{dx}} \sec p$.

The chain rule now becomes applicable as follows:

$\frac{d}{\mathrm{dx}} \sec p = \frac{d}{\mathrm{dp}} \sec p \cdot \frac{d}{\mathrm{dx}} p$

= $\sec p \tan p \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{\sqrt{x}}\right)$

= $\sec \left(\frac{1}{\sqrt{x}}\right) \tan \left(\frac{1}{\sqrt{x}}\right) \cdot \left(- \frac{1}{2} {x}^{- \frac{3}{2}}\right)$

=$\left(- \frac{1}{2}\right) \left(\frac{1}{x} ^ \left(\frac{3}{2}\right)\right) \sec \left(\frac{1}{\sqrt{x}}\right) \tan \left(\frac{1}{\sqrt{x}}\right)$