What is the derivative of # (tan(8x))^2#?

Question

2635 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-17T13:18:36

The derivative is #=16tan(8x)sec^2(8x)#

We need

#(u^n)'=n u^(n-1)* u'#

#(tanx)'=sec^2x#

We calculate this derivative by the chain rule

Let #y=(tan(8x))^2#

#dy/dx=2tan(8x)*sec^2(8x)*8#

#=16tan(8x)sec^2(8x)#