I don't know what I did wrong...
Find y' of y=cos^-1(e^2x)
Ok, so cos^-1 is sec.
Then I got sec(e^2x)tan(e^2x)(2e^2x)(2).
Not sure if the (2e^2x)(2) part is right because the online hw keep telling me I'm wrong.
You want to find the derivative of $\displaystyle y=cos^{-1}(e^{2x})$?.
Chain rule.
The derivative of $\displaystyle cos^{-1}(x)=\frac{-1}{\sqrt{1-x^{2}}}$
So, the derivative of $\displaystyle cos^{-1}(e^{2x})=\frac{-1}{\sqrt{1-e^{4x}}}\cdot{\underbrace{2e^{2x}}_{\text{chain rule}}}$
Therefore, $\displaystyle y'=\frac{-2e^{2x}}{\sqrt{1-e^{4x}}}$