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 stepbystep 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
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 stepbystep 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
Attachments

11.1 KB Views: 5