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

##### 1 Answer
Oct 19, 2017

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

#### Explanation:

The equation given can be simplified as $f \left(x\right) = \sec \left({\left(3 x\right)}^{- \frac{1}{2}}\right)$

If $f \left(x\right) = \sec \left(g \left(x\right)\right)$, then $f ' \left(x\right) = g ' \left(x\right) \sec \left(g \left(x\right)\right) \tan \left(g \left(x\right)\right)$

If $g \left(x\right) = {\left(h \left(x\right)\right)}^{n}$, then $g ' \left(x\right) = h \left(x\right) \cdot n {\left(h \left(x\right)\right)}^{n - 1}$

The derivative of $3 x$ is$3$.

$- \frac{1}{2} - 1 = - \frac{3}{2}$

$g ' \left(x\right) = 3 \cdot - \frac{1}{2} {\left(3 x\right)}^{- \frac{3}{2}} = - \frac{3 {\left(3 x\right)}^{- \frac{3}{2}}}{2}$

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