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

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Is a a constant?
are you differentiating with respect to x..?
or a? :P .

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Other answers:

Yes @campbell_st. I'm not sure @Dido525
That means a is a constant I assume O_o .
\(\ \text{ I got to this point, but my text tells me this answer is incorrect: } y'=-sinx(3a^2+3x^2). \)
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
I can't see the rest.
One moment
\(\ \Huge -sinx(3a^2+3x^2) \)
then replace u with a^3 + x^3 for the final solution.
nope its dy/dx = -sin(a^3 + x^3) * 3x^2
So the way my math teacher did this in class was using @Dido525 illustrated
@campbell_st How did you get that?
Because a is a constant. The derivative of a constant is 0.
We multiply by the derivative of the inside part when using the chain rule.
and I used the chain rule..
@camp is correct per calculus instructor
the a^3 x^3 doesn't change
@campbell_st I thought a was a constant then? How does that stay the same?
Here: |dw:1352349668500:dw|
We DO NOT change the inside.
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
Differentiate that.
With respect to x. Meaning a is a constant. We know the derivative of a constant is 0.
Usually letters a, b, and c are considered constants.
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?
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
Well there are partial derivatives but you probably won't learn that now.
@aerokat25 The correct answer the book provides is \(\ y'=-3x^2sin(a^3+x^3) .\) Why can't I differentiate x^3 ?
its just like (x^2 + 3)^4 dy/dx = 4(x^2 + 3) ^3 *2x the brackets don't change
you do... thats how you get 3x^2
When you have a function Just inside another function the function inside the brackets DO NOT change.
But the term is still inside the parenthesis? That's really confusing me right now
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
Chain rule says d/dx(cos u)= -sin u (u')
yep and du/dx = u' you are differentiating u with respect to x
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.
So u= a^3 + x^3
U'= 0 + 3x^2
well its the derivative of the outer function cos times the derivative of the inner function a^3 + x^3
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.
Ok well u know d/dx of cos x= -sin x right?
The expression inside of the expression on the outside ALWAYS remains. It's just how it is.
\(\ \text{Yes!} \)
"on the outside"?
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.
Okay\(\ \text{...} \)
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
Angle measure?

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