Thread: find derivative of a trig function (chain rule included)

Just trying to verify my work.
Find dy/dx for y=sin^5(cos(cos(3x)))

dy/dx= 5(sin(cos(cos(3x))))^4 * cos(cos(3x)) * (-sin(3x)) * 3

Does this look right to you all?

I get

I don't follow this answer at all, Pickslides. Can you explain the steps you followed????

dy/dx of y=sin^5(cos(cos3x)) which is also (sin(cos(cos(3x)))^5

Logically, we multiply the entire function by the exponent (5) and reduce it (the exponent) by 1.

y=5(sin^4(cos(cos(3x)))

Then we move along to the second layer (sin(cos(cos(3x))) which gives us

y=5(sin^4(cos(cos(3x))) * cos(cos(3x))

Then the third layer (cos(3x))

y=5(sin^4(cos(cos(3x))) * cos(cos(3x)) * (-sin(3x))

Then the fourth layer (3x)

y=5(sin^4(cos(cos(3x))) * cos(cos(3x)) * (-sin(3x)) * 3

It would help me greatly if you could point out the specific flaws in this logic. Thanks so much!

Take a look at this.

Be sure to ask it to show steps.

So now we have 3 answers... hahaha. And no explanations. Sad-face.jpg

Originally Posted by AlphaTauLambda

So now we have 3 answers... hahaha. And no explanations. Sad-face.jpg

But you have only one correct answer.

find dy/dx of sin^5(cos(3x))

I have, 5sin^4(cos(3x)) * cos(cos(3x)) * (-sin3x) * 3

Moderator edit: User made this correction to the original question that was asked.

Originally Posted by AlphaTauLambda

find dy/dx of sin^5(cos(3x))

I have, 5sin^4(cos(3x)) * cos(cos(3x)) * (-sin3x) * 3

Which looks good to me!

Originally Posted by Chris L T521

Which looks good to me!

It may look good to you, but it is wrong.
This is the correct answer.

Originally Posted by Plato

It may look good to you, but it is wrong.
This is the correct answer.

I agree that's the correct answer to what the user originally posted. However, they created a new thread with a "corrected" version (which I happened to merge into this one). So my comment was directed to the post which had the "fixed" version of the question.