# How do you differentiate f(x)=(sec^5 (1/x))^(1/3) using the chain rule?

##### 1 Answer
May 14, 2016

$\frac{d}{\mathrm{dx}} {\left({\sec}^{5} \left(\frac{1}{x}\right)\right)}^{\frac{1}{3}} = - \frac{5}{3 {x}^{2}} {\sec}^{\frac{5}{3}} \left(\frac{1}{x}\right) \tan \left(\frac{1}{x}\right)$

#### Explanation:

$f \left(x\right) = {\left({\sec}^{5} \left(\frac{1}{x}\right)\right)}^{\frac{1}{3}} = {\sec}^{\frac{5}{3}} \left(\frac{1}{x}\right)$

Here, we have $f \left(x\right) = g {\left(x\right)}^{\frac{5}{3}}$, $g \left(x\right) = \sec \left(h \left(x\right)\right)$ and $h \left(x\right) = \frac{1}{x}$

Hence according to chin rule

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}} \times \frac{\mathrm{dh}}{\mathrm{dx}}$

= $\frac{5}{3} {\left(g \left(x\right)\right)}^{\frac{2}{3}} \times \sec \left(h \left(x\right)\right) \tan \left(h \left(x\right)\right) \times \frac{- 1}{x} ^ 2$

= $\frac{5}{3} {\sec}^{\frac{2}{3}} \left(\frac{1}{x}\right) \times \sec \left(\frac{1}{x}\right) \tan \left(\frac{1}{x}\right) \times \frac{- 1}{x} ^ 2$

= $- \frac{5}{3 {x}^{2}} {\sec}^{\frac{5}{3}} \left(\frac{1}{x}\right) \tan \left(\frac{1}{x}\right)$