# How do you find the derivative of tan^2(3x)?

Apr 20, 2015

In this way with the chain rule:

$y ' = 2 \tan 3 x \cdot \frac{1}{\cos} ^ 2 \left(3 x\right) \cdot 3 = \frac{6 \tan 3 x}{\cos} ^ 2 \left(3 x\right)$

or

$y ' = 2 \tan 3 x \cdot \left(1 + {\tan}^{2} 3 x\right) \cdot 3 = 6 \tan 3 x \left(1 + {\tan}^{2} 3 x\right)$,

and this is because $\frac{1}{\cos} {x}^{2} \left(3 x\right) = 1 + {\tan}^{2} 3 x$.

Apr 20, 2015

I will be solving the problem using implicit differentiation...