{Derivatives, Focus: Chain Rule. Help appreciated}
\(\ \Huge y=cos(a^3+x^3) . \)

- anonymous

{Derivatives, Focus: Chain Rule. Help appreciated}
\(\ \Huge y=cos(a^3+x^3) . \)

- chestercat

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- anonymous

Is a a constant?

- campbell_st

are you differentiating with respect to x..?

- anonymous

or a? :P .

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## More answers

- anonymous

Yes @campbell_st. I'm not sure @Dido525

- anonymous

That means a is a constant I assume O_o .

- anonymous

\(\
\text{ I got to this point, but my text tells me this answer is incorrect: }
y'=-sinx(3a^2+3x^2).
\)

- anonymous

|dw:1352349392745:dw|

- campbell_st

let u = a^3 + x^3 du/dx = 3x^2
let y = cos(u) du/dy = -sin (u)
so dy/dx = dy/du * du/dx

- anonymous

I can't see the rest.

- anonymous

One moment

- anonymous

\(\ \Huge -sinx(3a^2+3x^2) \)

- campbell_st

then replace u with a^3 + x^3 for the final solution.

- campbell_st

nope its
dy/dx = -sin(a^3 + x^3) * 3x^2

- anonymous

So the way my math teacher did this in class was using @Dido525 illustrated

- anonymous

@campbell_st How did you get that?

- anonymous

Because a is a constant. The derivative of a constant is 0.

- anonymous

We multiply by the derivative of the inside part when using the chain rule.

- campbell_st

and I used the chain rule..

- anonymous

@camp is correct per calculus instructor

- campbell_st

the a^3 x^3 doesn't change

- anonymous

@campbell_st I thought a was a constant then? How does that stay the same?

- anonymous

Here:
|dw:1352349668500:dw|

- anonymous

We DO NOT change the inside.

- anonymous

Now if you are implicitly differentiating then you could possibly get a different answer if a is not constant. However I'm pretty sure a is a constant in this case

- anonymous

|dw:1352349758941:dw|

- anonymous

Differentiate that.

- anonymous

With respect to x. Meaning a is a constant. We know the derivative of a constant is 0.

- anonymous

Usually letters a, b, and c are considered constants.

- anonymous

So you can't differentiate more than one variable in any given function? Only variable is differentiated in a problem with more than one variable?

- anonymous

Also check to see if you typed the question correctly because that answer you typed from the back of the book looks like you changed the angle measurement to x

- anonymous

Well there are partial derivatives but you probably won't learn that now.

- anonymous

@aerokat25 The correct answer the book provides is \(\ y'=-3x^2sin(a^3+x^3) .\) Why can't I differentiate x^3 ?

- campbell_st

its just like
(x^2 + 3)^4 dy/dx = 4(x^2 + 3) ^3 *2x
the brackets don't change

- campbell_st

you do... thats how you get 3x^2

- anonymous

When you have a function Just inside another function the function inside the brackets DO NOT change.

- anonymous

But the term is still inside the parenthesis? That's really confusing me right now

- anonymous

parentheses*

- campbell_st

that is why textbooks talk about
u = a^3 + x^3 du/dx = 3x^2
y = cos(u) dy/du = -sin(u)
dy/dx = dy/du * du/dx

- anonymous

|dw:1352350103544:dw|

- anonymous

Chain rule says d/dx(cos u)= -sin u (u')

- campbell_st

yep and du/dx = u'
you are differentiating u with respect to x

- anonymous

Hmmm. That's really confusing because our teacher shows us the u method on one example, and said he doesn't use it. So I don't really follow, to be honest.

- anonymous

So u= a^3 + x^3

- anonymous

U'= 0 + 3x^2

- campbell_st

well its the derivative of the outer function cos times the derivative of the inner function a^3 + x^3

- anonymous

Okay, I get that part, but the one thing I am not understanding is how the expression remains. I'm sorry if I'm being silly or something.

- anonymous

Ok well u know d/dx of cos x= -sin x right?

- anonymous

The expression inside of the expression on the outside ALWAYS remains. It's just how it is.

- anonymous

\(\ \text{Yes!} \)

- anonymous

"on the outside"?

- anonymous

Well d/dx cos x actually equals -sin x (x') but the derivative of x is just 1 so you have actually been using chain all along and didn't really know it.

- anonymous

Okay\(\ \text{...} \)

- anonymous

And that angle measure x has never changed right? So you aren't going to change what you're given as an angle measurement now either. You just have to multiply by its derivative which isn't going to be a 1 anymore

- anonymous

Angle measure?

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