anonymous
  • anonymous
{Derivatives, Focus: Chain Rule. Help appreciated} \(\ \Huge y=cos(a^3+x^3) . \)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Is a a constant?
campbell_st
  • campbell_st
are you differentiating with respect to x..?
anonymous
  • anonymous
or a? :P .

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anonymous
  • anonymous
Yes @campbell_st. I'm not sure @Dido525
anonymous
  • anonymous
That means a is a constant I assume O_o .
anonymous
  • anonymous
\(\ \text{ I got to this point, but my text tells me this answer is incorrect: } y'=-sinx(3a^2+3x^2). \)
anonymous
  • anonymous
|dw:1352349392745:dw|
campbell_st
  • 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
  • anonymous
I can't see the rest.
anonymous
  • anonymous
One moment
anonymous
  • anonymous
\(\ \Huge -sinx(3a^2+3x^2) \)
campbell_st
  • campbell_st
then replace u with a^3 + x^3 for the final solution.
campbell_st
  • campbell_st
nope its dy/dx = -sin(a^3 + x^3) * 3x^2
anonymous
  • anonymous
So the way my math teacher did this in class was using @Dido525 illustrated
anonymous
  • anonymous
@campbell_st How did you get that?
anonymous
  • anonymous
Because a is a constant. The derivative of a constant is 0.
anonymous
  • anonymous
We multiply by the derivative of the inside part when using the chain rule.
campbell_st
  • campbell_st
and I used the chain rule..
anonymous
  • anonymous
@camp is correct per calculus instructor
campbell_st
  • campbell_st
the a^3 x^3 doesn't change
anonymous
  • anonymous
@campbell_st I thought a was a constant then? How does that stay the same?
anonymous
  • anonymous
Here: |dw:1352349668500:dw|
anonymous
  • anonymous
We DO NOT change the inside.
anonymous
  • 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
  • anonymous
|dw:1352349758941:dw|
anonymous
  • anonymous
Differentiate that.
anonymous
  • anonymous
With respect to x. Meaning a is a constant. We know the derivative of a constant is 0.
anonymous
  • anonymous
Usually letters a, b, and c are considered constants.
anonymous
  • 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
  • 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
  • anonymous
Well there are partial derivatives but you probably won't learn that now.
anonymous
  • 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
  • campbell_st
its just like (x^2 + 3)^4 dy/dx = 4(x^2 + 3) ^3 *2x the brackets don't change
campbell_st
  • campbell_st
you do... thats how you get 3x^2
anonymous
  • anonymous
When you have a function Just inside another function the function inside the brackets DO NOT change.
anonymous
  • anonymous
But the term is still inside the parenthesis? That's really confusing me right now
anonymous
  • anonymous
parentheses*
campbell_st
  • 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
  • anonymous
|dw:1352350103544:dw|
anonymous
  • anonymous
Chain rule says d/dx(cos u)= -sin u (u')
campbell_st
  • campbell_st
yep and du/dx = u' you are differentiating u with respect to x
anonymous
  • 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
  • anonymous
So u= a^3 + x^3
anonymous
  • anonymous
U'= 0 + 3x^2
campbell_st
  • campbell_st
well its the derivative of the outer function cos times the derivative of the inner function a^3 + x^3
anonymous
  • 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
  • anonymous
Ok well u know d/dx of cos x= -sin x right?
anonymous
  • anonymous
The expression inside of the expression on the outside ALWAYS remains. It's just how it is.
anonymous
  • anonymous
\(\ \text{Yes!} \)
anonymous
  • anonymous
"on the outside"?
anonymous
  • 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
  • anonymous
Okay\(\ \text{...} \)
anonymous
  • 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
  • anonymous
Angle measure?

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