# If f(x) =tan^2x  and g(x) = sqrt(5x-1 , what is f'(g(x)) ?

Aug 2, 2017

$f ' \left(g \left(x\right)\right) = \frac{5 {\sec}^{2} \left(\sqrt{5 x - 1}\right) \tan \left(\sqrt{5 x - 1}\right)}{\sqrt{5 x - 1}}$

#### Explanation:

To find $f ' \left(g \left(x\right)\right)$, we can start by finding $f \left(g \left(x\right)\right)$ and then take the derivative. We substitute $\sqrt{5 x - 1}$ into $f \left(x\right)$, so we have:

$f \left(g \left(x\right)\right) = f \left(\sqrt{5 x - 1}\right) = {\tan}^{2} \left(\sqrt{5 x - 1}\right)$

Now we take the derivative using the chain rule.

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

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