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Study23
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{Derivatives, Focus: Chain Rule. Help appreciated}
\(\ \Huge y=cos(a^3+x^3) . \)
 2 years ago
 2 years ago
Study23 Group Title
{Derivatives, Focus: Chain Rule. Help appreciated} \(\ \Huge y=cos(a^3+x^3) . \)
 2 years ago
 2 years ago

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Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Is a a constant?
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
are you differentiating with respect to x..?
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Yes @campbell_st. I'm not sure @Dido525
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
That means a is a constant I assume O_o .
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
\(\ \text{ I got to this point, but my text tells me this answer is incorrect: } y'=sinx(3a^2+3x^2). \)
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
dw:1352349392745:dw
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
I can't see the rest.
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
\(\ \Huge sinx(3a^2+3x^2) \)
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
then replace u with a^3 + x^3 for the final solution.
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
nope its dy/dx = sin(a^3 + x^3) * 3x^2
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
So the way my math teacher did this in class was using @Dido525 illustrated
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
@campbell_st How did you get that?
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Because a is a constant. The derivative of a constant is 0.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
We multiply by the derivative of the inside part when using the chain rule.
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
and I used the chain rule..
 2 years ago

aerokat25 Group TitleBest ResponseYou've already chosen the best response.0
@camp is correct per calculus instructor
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
the a^3 x^3 doesn't change
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
@campbell_st I thought a was a constant then? How does that stay the same?
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Here: dw:1352349668500:dw
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
We DO NOT change the inside.
 2 years ago

aerokat25 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
dw:1352349758941:dw
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Differentiate that.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
With respect to x. Meaning a is a constant. We know the derivative of a constant is 0.
 2 years ago

thewilled1 Group TitleBest ResponseYou've already chosen the best response.0
Usually letters a, b, and c are considered constants.
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

aerokat25 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
Well there are partial derivatives but you probably won't learn that now.
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
@aerokat25 The correct answer the book provides is \(\ y'=3x^2sin(a^3+x^3) .\) Why can't I differentiate x^3 ?
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
its just like (x^2 + 3)^4 dy/dx = 4(x^2 + 3) ^3 *2x the brackets don't change
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
you do... thats how you get 3x^2
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
When you have a function Just inside another function the function inside the brackets DO NOT change.
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
But the term is still inside the parenthesis? That's really confusing me right now
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
parentheses*
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
dw:1352350103544:dw
 2 years ago

aerokat25 Group TitleBest ResponseYou've already chosen the best response.0
Chain rule says d/dx(cos u)= sin u (u')
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
yep and du/dx = u' you are differentiating u with respect to x
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

aerokat25 Group TitleBest ResponseYou've already chosen the best response.0
So u= a^3 + x^3
 2 years ago

aerokat25 Group TitleBest ResponseYou've already chosen the best response.0
U'= 0 + 3x^2
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
well its the derivative of the outer function cos times the derivative of the inner function a^3 + x^3
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

aerokat25 Group TitleBest ResponseYou've already chosen the best response.0
Ok well u know d/dx of cos x= sin x right?
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.0
The expression inside of the expression on the outside ALWAYS remains. It's just how it is.
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
\(\ \text{Yes!} \)
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
"on the outside"?
 2 years ago

aerokat25 Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Okay\(\ \text{...} \)
 2 years ago

aerokat25 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Angle measure?
 2 years ago
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