How do you find the derivative of #y = sin(tan 2x)#?

1 Answer
Nov 13, 2015

#(2cos(tan(2x)))/(cos^2(2x))#

Explanation:

You must use the chain rule: this means that

#(f(g(h(x)))' = f'(g(h(x))) * g'(h(x)) * h'(x)#

In your case, we have:

  • #f(x)=sin(x)#, and thus #f'(x)=cos(x)#;
  • #g(x)=tan(x)#, and thus #g'(x)=1/cos^2(x)#;
  • #h(x)=2x#, and thus #h'(x)=2#.

Plugging these functions into the original formula gives:

  • #f'(g(h(x))) = cos(tan(2x))#
  • #g'(h(x)) = 1/(cos^2(2x))#
  • #h'(x) = 2#

Multiplying the three, you get

#(2cos(tan(2x)))/(cos^2(2x))#