Finding the second derivative of trig function

Mar 2017
356
3
Massachusetts
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
 

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ChipB

MHF Helper
Jun 2014
310
128
NJ
Your result and the book's answer for the first derivative are the same thing! It almost always help to work in terms of sines and cosines rather than secants and tangents. Your answer: y' = cosx tanx + sinx sec^2x = cos x (sinx/cosx) + sinx/(cos^2x) = sinx + sinx/(cos^2x) = sinx + (1/cosx) (sinx/cosx) = sinx + secx tanx = their answer.

As for the second derivative: start with y' = sinx + sinx/(cos^2x) and go from there.
 
Last edited:
Mar 2017
356
3
Massachusetts
ahhh okay makes sense now... thank you