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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:

Set:

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

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

\(w(t)=t^5\)

The function can be rewritten using these three function as:

\(f(t)=w(v(u(t)))\)

\(\frac{d}{dt}\left[w\left(v\left(u\left(t\right)\right)\right)\right]=w'\left(v\left(u\left(t\right)\right)\right).v'\left(u\left(t\right)\right).u'\left(t\right)\)

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 \(e^{t^2}\)

\(x(t)=t^2\)

\(y(t)=e^t\)

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

\(y(x(t))\)

Apply regular chain rule:

\(\frac{d}{dt}\left[y\left(x\left(t\right)\right)\right]=y'\left(x\left(t\right)\right).x'\left(t\right)\)

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

\(\frac{d}{dt}\left[e^{t^2}\right]=e^{t^2}.2t=2te^{t^2}\)

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

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

Jeffes02 Aug 7, 2017

#1**+3 **

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

https://www.wolframalpha.com/input/?i=y%3D%5Bsin(e%5E(t%5E2)-cos(t))%5D%5E5++differentiate

.Melody Aug 7, 2017

#2**+1 **

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

Jeffes02 Aug 7, 2017