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

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campbell_stBest ResponseYou've already chosen the best response.1
are you differentiating with respect to x..?
 one year ago

Study23Best ResponseYou've already chosen the best response.0
Yes @campbell_st. I'm not sure @Dido525
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
That means a is a constant I assume O_o .
 one year ago

Study23Best ResponseYou've already chosen the best response.0
\(\ \text{ I got to this point, but my text tells me this answer is incorrect: } y'=sinx(3a^2+3x^2). \)
 one year ago

campbell_stBest 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
 one year ago

Study23Best ResponseYou've already chosen the best response.0
\(\ \Huge sinx(3a^2+3x^2) \)
 one year ago

campbell_stBest ResponseYou've already chosen the best response.1
then replace u with a^3 + x^3 for the final solution.
 one year ago

campbell_stBest ResponseYou've already chosen the best response.1
nope its dy/dx = sin(a^3 + x^3) * 3x^2
 one year ago

Study23Best ResponseYou've already chosen the best response.0
So the way my math teacher did this in class was using @Dido525 illustrated
 one year ago

Study23Best ResponseYou've already chosen the best response.0
@campbell_st How did you get that?
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Because a is a constant. The derivative of a constant is 0.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
We multiply by the derivative of the inside part when using the chain rule.
 one year ago

campbell_stBest ResponseYou've already chosen the best response.1
and I used the chain rule..
 one year ago

aerokat25Best ResponseYou've already chosen the best response.0
@camp is correct per calculus instructor
 one year ago

campbell_stBest ResponseYou've already chosen the best response.1
the a^3 x^3 doesn't change
 one year ago

Study23Best ResponseYou've already chosen the best response.0
@campbell_st I thought a was a constant then? How does that stay the same?
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Here: dw:1352349668500:dw
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
We DO NOT change the inside.
 one year ago

aerokat25Best 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
 one year ago

Dido525Best 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.
 one year ago

thewilled1Best ResponseYou've already chosen the best response.0
Usually letters a, b, and c are considered constants.
 one year ago

Study23Best 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?
 one year ago

aerokat25Best 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
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Well there are partial derivatives but you probably won't learn that now.
 one year ago

Study23Best 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 ?
 one year ago

campbell_stBest 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
 one year ago

campbell_stBest ResponseYou've already chosen the best response.1
you do... thats how you get 3x^2
 one year ago

Dido525Best 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.
 one year ago

Study23Best ResponseYou've already chosen the best response.0
But the term is still inside the parenthesis? That's really confusing me right now
 one year ago

campbell_stBest 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
 one year ago

aerokat25Best ResponseYou've already chosen the best response.0
Chain rule says d/dx(cos u)= sin u (u')
 one year ago

campbell_stBest ResponseYou've already chosen the best response.1
yep and du/dx = u' you are differentiating u with respect to x
 one year ago

Study23Best 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.
 one year ago

campbell_stBest 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
 one year ago

Study23Best 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.
 one year ago

aerokat25Best ResponseYou've already chosen the best response.0
Ok well u know d/dx of cos x= sin x right?
 one year ago

Dido525Best 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.
 one year ago

aerokat25Best 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.
 one year ago

aerokat25Best 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
 one year ago
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