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Thread: Finding the second derivative of trig function

  1. #1
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    Question Finding the second derivative of trig function

    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
    Attached Thumbnails Attached Thumbnails Finding the second derivative of trig function-screen-shot-2017-09-11-10.13.18-am.png  
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  2. #2
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    Re: Finding the second derivative of trig function

    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 by ChipB; Sep 11th 2017 at 06:59 AM.
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  3. #3
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    Re: Finding the second derivative of trig function

    ahhh okay makes sense now... thank you
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