1. ## Cosine Function Derivative

Question # 2 lol. I don't know why I'm having such a hard time getting this....

Determine the derivative of the following fuction:
g(x) = (cosx)^3

I started by using the power rule to come up with:

g '(x) = 3(cosx)^2
= 3(-sinx)^2

According to my book this is wrong though... it has:

g '(x) = (3(cosx)^2)(-sinx)
= -3cos^2xsinx

Which makes no sense to me. Thanks again.

2. $\displaystyle y= \cos ^3x$

using the chain rule $\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

so make $\displaystyle u = \cos x$ so $\displaystyle y = u^3$

now

$\displaystyle \frac{dy}{du}= 3u^2$

and

$\displaystyle \frac{du}{dx}= -\sin x$

$\displaystyle \frac{dy}{dx} = 3u^2 \times -\sin x= 3\cos^2x \times -\sin x = -3\cos^2x\sin x$

3. I was taught the chain rule a little differently. It was shown to me like this:

f (x) = a^g(x)
f '(x) = (a^g(x))(ln a)(g '(x))

Would it be possible to quickly go over how the method you are using works:

dy/dx = (dy/du)(du/dx)

Thanks!

4. Originally Posted by Jools

Would it be possible to quickly go over how the method you are using works:

dy/dx = (dy/du)(du/dx)

Thanks!
This is exactly what I have done above. This is the chain rule.

Originally Posted by Jools
I was taught the chain rule a little differently. It was shown to me like this:

f (x) = a^g(x)
f '(x) = (a^g(x))(ln a)(g '(x))
This is just a version of the chain rule, it won't work in your case because your function is not in the correct form to use this adaptation.

5. Sorry, I'm confused. Let's take it from the top. First off how did you come up with y = cos^3 x? Then I think I just need to know what each variable in the formula represents and I should be able to figure this out. I appreciate your patience.

6. Originally Posted by Jools
I was taught the chain rule a little differently. It was shown to me like this:

f (x) = a^g(x)
f '(x) = (a^g(x))(ln a)(g '(x))
This technique only works if a is a constant.

I find it easiest to remember this:

if f(x) =[g(x)]^a, then
f'(x) = a*[g(x)]^(a-1)*g'(x)

Here's an examples:

f(x) = (3x)^3
f'(x) = 3*(3x)^2 *3 = 81x^2

Another example:

f(x) = (x^2)^3
f'(x) = 3*(x^2)^2*2x = 6x^5

See how it works? So applying it to your problem:

f(x) = [cos(x)]^3
f'(x) = 3* [cos(x)]^2 * (-sin(x))
= -3 cos^2(x) sin(x)

7. Originally Posted by Jools
Sorry, I'm confused. Let's take it from the top. First off how did you come up with y = cos^3 x? Then I think I just need to know what each variable in the formula represents and I should be able to figure this out. I appreciate your patience.
Ah yes ... the notation cos^3(x) means the the same thing as [cos(x)]^3. Perhaps that's the confusion.

8. So how exactly do I determine what the g(x) is? For example taking another question asking for the derivative:

f(x) = sin(x^2)

Is x^2 the g(x)?

9. Originally Posted by Jools
So how exactly do I determine what the g(x) is? For example taking another question asking for the derivative:

f(x) = sin(x^2)

Is x^2 the g(x)?
Yes, that is correct.

Here is the chain rule written in terms of the composite function:

$\displaystyle \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

10. Originally Posted by icemanfan
Yes, that is correct.

Here is the chain rule written in terms of the composite function:

$\displaystyle \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$
Ok I think I'm catching on... So one more...

g(x) = e^cosx

How would that one go?

11. Originally Posted by Jools
Ok I think I'm catching on... So one more...

g(x) = e^cosx

How would that one go?
Let $\displaystyle u = \cos{x}$ so that $\displaystyle g = e^u$.

$\displaystyle \frac{du}{dx} = -\sin{x}$

$\displaystyle \frac{dg}{du} = e^u = e^{\cos{x}}$

So $\displaystyle \frac{dg}{dx} = -\sin{x}\,e^{\cos{x}}$.

12. Originally Posted by Prove It
Let $\displaystyle u = \cos{x}$ so that $\displaystyle g = e^u$.

$\displaystyle \frac{du}{dx} = -\sin{x}$

$\displaystyle \frac{dg}{du} = e^u = e^{\cos{x}}$

So $\displaystyle \frac{dg}{dx} = -\sin{x}\,e^{\cos{x}}$.
Could that be applied to this formula?

f '(g(x))(g '(x))

13. Originally Posted by Jools
Could that be applied to the formula?

f '(g(x))(g '(x))
Yes - just replace where I've used $\displaystyle u$ with $\displaystyle g(x)$ and replace where I've used $\displaystyle g$ with $\displaystyle f$.

14. Originally Posted by Prove It
Yes - just replace where I've used $\displaystyle u$ with $\displaystyle g(x)$ and replace where I've used $\displaystyle g$ with $\displaystyle f$.
Got it. I REALLY appreciate all the help. I understand how to work out the derivative I think... I'm just having a hard time determining what the value of g(x) is for each variation....