# How do you differentiate f(x)=(tan^2 3x)^(7/3) using the chain rule?

Apr 5, 2016

$\frac{42}{3} {\left({\tan}^{2} 3 x\right)}^{\frac{4}{3}} \tan 3 x {\sec}^{2} 3 x$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{ chain rule }}$

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) . g ' \left(x\right)$
$\text{------------------------------------------------------------------------}$

f(g(x)) =${\left({\tan}^{2} 3 x\right)}^{\frac{7}{3}} \Rightarrow f ' \left(g \left(x\right)\right) = \frac{7}{3} {\left({\tan}^{2} 3 x\right)}^{\frac{4}{3}}$

g(x) = ${\tan}^{2} 3 x \Rightarrow g ' \left(x\right) = 2 \tan 3 x . \frac{d}{\mathrm{dx}} \left(\tan 3 x\right)$

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

Now combining these results together

f'(x) = $\frac{7}{3} {\left({\tan}^{2} 3 x\right)}^{\frac{4}{3}} .2 \tan 3 x .3 {\sec}^{2} 3 x$

$= \frac{42}{3} {\left({\tan}^{2} 3 x\right)}^{\frac{4}{3}} \tan 3 x {\sec}^{2} 3 x$