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

Mar 16, 2016

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

#### Explanation:

As $f \left(x\right) = {\tan}^{2} \left(\frac{x}{2}\right)$,

$f ' \left(x\right) = \frac{d \left({\tan}^{2} \left(\frac{x}{2}\right)\right)}{d \left(\tan \frac{x}{2}\right)}$$\times$$\frac{d \left(\tan \frac{x}{2}\right)}{d \left(\frac{x}{2}\right)}$$\times$$\frac{d \left(\frac{x}{2}\right)}{\mathrm{dx}}$

Hence $f ' \left(x\right) = 2 \tan \left(\frac{x}{2}\right) \times {\sec}^{2} \left(\frac{x}{2}\right) \times \frac{1}{2}$

or $f ' \left(x\right) = \tan \left(\frac{x}{2}\right) \times {\sec}^{2} \left(\frac{x}{2}\right)$

Hence $f ' \left(g \left(x\right)\right) = \tan \left(g \frac{x}{2}\right) \times {\sec}^{2} \left(g \frac{x}{2}\right)$

and as $g \left(x\right) = \sqrt{5 x - 1}$

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