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

1. ## 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?

2. I get

$\displaystyle \displaystyle \frac{dy}{dx} = 5\sin^4(\cos (\cos 3x)) \times 3\cos (\cos 3x) \times \sin 3x \times \sin (\cos 3x)$

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

4. ## 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!

5. Take a look at this.

Be sure to ask it to show steps.

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

7. Originally Posted by AlphaTauLambda
But you have only one correct answer.

8. ## 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.

9. 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!

10. Originally Posted by Chris L T521
Which looks good to me!
It may look good to you, but it is wrong.