1. ## deriv again

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.

2. Mmm... be careful with the notation.

$\displaystyle \cos^{-1}x=\arccos x$

--

You wanna compute $\displaystyle y=\sec e^{2x}$

You know that $\displaystyle (\sec u)'=u'\cdot\sec u\tan u$

3. So then I got sec(e^2x)tan(e^2x)(2e^2x).

I'm still not getting it right.

4. 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, $\displaystyle y'=\frac{-2e^{2x}}{\sqrt{1-e^{4x}}}$