Hi,

I hope someone can help. I'm trying to find the second derivative for the trigonometric function y = sinxtanx

Apparently the first derivative is y' = sinx + secxtanx even though I got y' = cosxtanx + sinx(secx)^2. I must of missed a step and would appreciate help.

The second derivative is y'' = cosx + secx + (2sinx)^2/(cosx)^3. Showing a step-by-step on how to get to this as well would be wonderful...

While I understand that there are various ways to express the first and second derivative, I would like someone to explain how to get to the expressions which I just mentioned.

- Olivia