Question

How do you take the derivative of `tan^3 (3x-1)`?

Answer 1

`9tan^2(3x-1)sec^2(3x-1)`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`"here "y=tan^3(3x-1)=(tan(3x-1))^3`

`rArrdy/dx=3(tan(3x-1))^2xxd/dx(tan(3x-1))`

`color(white)(rArrdy/dx)=3tan^2(3x-1)xxsec^2(3x-1)xxd/dx(3x-1)`

`color(white)(rArrdy/dx)=9tan^2(3x-1)sec^2(3x-1)`