# How do you differentiate f(x)=sec(1/(5x^2-1)) using the chain rule?

Jun 24, 2016

Let $y = \sec u$

and $u = {\left(5 {x}^{2} - 1\right)}^{- 1}$

Now, we have to differentiate $y$ and $u$. We will have to use the chain rule to differentiate u.

$u = {v}^{-} 1$

$v = 5 {x}^{2} - 1$

$\frac{d}{\mathrm{dx}} \left({\left(5 {x}^{2} - 1\right)}^{-} 1\right) = 10 x \times - \frac{1}{{v}^{2}}$

$= - \frac{10 x}{v} ^ 2$

$= - \frac{10 x}{5 {x}^{2} - 1} ^ 2$

Hence, $u ' = - \frac{10 x}{5 {x}^{2} - 1} ^ 2$.

The derivative of $\sec u$ is $- \csc u$, because $\sec u = \frac{1}{\cos} u$ and $\left(\frac{1}{\cos} u\right) ' = - \frac{1}{\sin} u = - \csc u$.

$f ' \left(x\right) = - \csc u \times - \frac{10 x}{5 {x}^{2} - 1} ^ 2$

$f ' \left(x\right) = \frac{10 x \csc \left(\frac{1}{5 {x}^{2} - 1}\right)}{5 {x}^{2} - 1} ^ 2$

Hopefully this helps!