I was on the internet randomly searching for derivative worksheets to practice AP Calculus AB when I came across this:

\(f\left(t\right)=\sin ^5\left(e^{t^2}-\cos \left(t\right)\right)\)

I used the triple composite function of chain rule two times to solve this:

\(\frac{d}{dt}\left[\sin ^5\left(e^{t^2}-\cos \left(t\right)\right)\right]\)

Apply advanced chain rule:


\(u\left(t\right)=e^{t^2}-\cos \left(t\right)\)

\(v\left(t\right)=\sin \left(t\right)\)


The function can be rewritten using these three function as:



Unsubstitute \(u(t), v(t), \)and \(w(t)\), I got:

\(\frac{d}{dt}\left[\sin ^5\left(e^{t^2}-\cos \left(t\right)\right)\right]\)

\(=4\left(\sin ^4\left(e^{t^2}-\cos \left(t\right)\right)\right).\cos \left(e^{t^2}-\cos \left(t\right)\right).\frac{d}{dx}\left[e^{t^2}-\cos \left(t\right)\right]\)

Applying chain rule for the second time on \(e^{t^2}\)



The function \(e^{t^2}\) can be rewritten as:


Apply regular chain rule:


Unsubstitute \(x(t)\) and \(y(t)\), I got


\(\frac{d}{dt}\left[-\cos \left(t\right)\right]=\sin \left(t\right)\)

Place the solved integral back:

\(\frac{d}{dt}\left[\sin ^5\left(e^{t^2}-\cos \left(t\right)\right)\right]\)

\(=4\left(\sin ^4\left(e^{t^2}-\cos \left(t\right)\right)\right).\cos \left(e^{t^2}-\cos \left(t\right)\right).\left(2te^{t^2}+\sin \left(t\right)\right)\)

Yet when I compare the correct answer to my answer, my answer seems to be constantly about 3/5 times the correct answer's value, I stared eye-to-eye to my process for three hours but can't find any problems, can I get anyone's help? :P

People who can help me will be highly-appreciated :D

 Aug 7, 2017
edited by Guest  Aug 7, 2017
edited by Guest  Aug 7, 2017
edited by Guest  Aug 7, 2017
edited by Jeffes02  Aug 7, 2017
edited by Jeffes02  Aug 7, 2017

The 4 out the front should be a 5. I assume this is a careless oversight on your part.


\(y=(sinu)^5\\ \frac{dy}{du}=5(sinu)^4\)


I have answered it myself and the rest of your answer is correct.  (I only compared our end end results.)


I also double checked with Wolfram Alpha and it agrees too



 Aug 7, 2017
edited by Melody  Aug 7, 2017

Oh yes I didn't notice that XD I'm so dumb, I appreciate you and your sharp eyes!

 Aug 7, 2017

You are far from dumb. :)

Sometimes when a person makes a trivial mistake like that it can be very difficult for the person to detect the error. And yes it is  can feel a little humiliating when the mistake is pointed out. But don't fret, we all do it from time to time :)

Melody  Aug 7, 2017

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