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

This Question is Closed

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

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

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

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

campbell_st
 2 years ago
Best ResponseYou've already chosen the best response.1let u = a^3 + x^3 du/dx = 3x^2 let y = cos(u) du/dy = sin (u) so dy/dx = dy/du * du/dx

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

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

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

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

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

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

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

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

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

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

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

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

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

aerokat25
 2 years ago
Best ResponseYou've already chosen the best response.0Now 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

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

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

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0So you can't differentiate more than one variable in any given function? Only variable is differentiated in a problem with more than one variable?

aerokat25
 2 years ago
Best ResponseYou've already chosen the best response.0Also 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

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

Study23
 2 years ago
Best 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 ?

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

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

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

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

campbell_st
 2 years ago
Best ResponseYou've already chosen the best response.1that 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

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

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

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0Hmmm. 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.

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

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, 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.

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

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

aerokat25
 2 years ago
Best ResponseYou've already chosen the best response.0Well 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.

aerokat25
 2 years ago
Best ResponseYou've already chosen the best response.0And 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
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