# find derivative of a trig function (chain rule included)

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• May 8th 2011, 02:14 PM
AlphaTauLambda
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?
• May 8th 2011, 02:27 PM
pickslides
I get

$\displaystyle \displaystyle \frac{dy}{dx} = 5\sin^4(\cos (\cos 3x)) \times 3\cos (\cos 3x) \times \sin 3x \times \sin (\cos 3x)$
• May 8th 2011, 02:39 PM
AlphaTauLambda
I don't follow this answer at all, Pickslides. Can you explain the steps you followed????
• May 8th 2011, 02:45 PM
AlphaTauLambda
reasoning
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!
• May 8th 2011, 03:13 PM
Plato
Take a look at this.

Be sure to ask it to show steps.
• May 8th 2011, 03:19 PM
AlphaTauLambda
So now we have 3 answers... hahaha. And no explanations. Sad-face.jpg
• May 8th 2011, 03:23 PM
Plato
Quote:

Originally Posted by AlphaTauLambda
So now we have 3 answers... hahaha. And no explanations. Sad-face.jpg

But you have only one correct answer.
• May 8th 2011, 04:10 PM
AlphaTauLambda
mistyped equation - trig function with chain rule
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.
• May 8th 2011, 04:16 PM
Chris L T521
Quote:

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! :)
• May 8th 2011, 05:06 PM
Plato
Quote:

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.
• May 8th 2011, 05:20 PM
Chris L T521
Quote:

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.