Question

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

Answer 1

`(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))`