Thread: dy/dx = dy/du * du/dx

1. dy/dx = dy/du * du/dx

Hey.

I was learning about how to differentiate composite functions. This is what I read:

"If y = c(ax + b) + d.
let u = ax + b, then y = cu + d.
So dy/du = c, du/dx = a, and dy/dx = ca, that is dy/dx = dy/du * du/dx

The equation dy/du = c means that y is changing c times as fast as u; similarly u is changing a times as fast as x. It is natural to think that if y is changing c times as fast as u, and u is changing a times as fast as x, then y is changing c*a times as fast as x. Thus again, dy/dx = dy/du * du/dx"

I'm not surprised that it wasn't natural for me to think that way, and even after thinking about it, I still don't understand. I just can't get my head around it! Could somebody explain

2. Originally Posted by burgerandcheese
Hey.

I was learning about how to differentiate composite functions. This is what I read:

"If y = c(ax + b) + d.
let u = ax + b, then y = cu + d.
So dy/du = c, du/dx = a, and dy/dx = ca, that is dy/dx = dy/du * du/dx

The equation dy/du = c means that y is changing c times as fast as u; similarly u is changing a times as fast as x. It is natural to think that if y is changing c times as fast as u, and u is changing a times as fast as x, then y is changing c*a times as fast as x. Thus again, dy/dx = dy/du * du/dx"

I'm not surprised that it wasn't natural for me to think that way, and even after thinking about it, I still don't understand. I just can't get my head around it! Could somebody explain
y = c(ax + b) + d = cax+cb+d. So dy/dx =ca

3. Originally Posted by Jomo
y = c(ax + b) + d = cax+cb+d. So dy/dx =ca
Well, yes of course! That's the easier way

4. Originally Posted by burgerandcheese
Hey.

I was learning about how to differentiate composite functions. This is what I read:

"If y = c(ax + b) + d.
let u = ax + b, then y = cu + d.
So dy/du = c, du/dx = a, and dy/dx = ca, that is dy/dx = dy/du * du/dx

The equation dy/du = c means that y is changing c times as fast as u; similarly u is changing a times as fast as x. It is natural to think that if y is changing c times as fast as u, and u is changing a times as fast as x, then y is changing c*a times as fast as x. Thus again, dy/dx = dy/du * du/dx"

I'm not surprised that it wasn't natural for me to think that way, and even after thinking about it, I still don't understand. I just can't get my head around it! Could somebody explain
Suppose I run twice as fast as Tim, and Tim runs 3 times as fast as Dan. Then I run 6 times as fast as Dan.

Is that the part you don't see, or is it how that relates to the derivative?

5. Originally Posted by burgerandcheese
Well, yes of course! That's the easier way
That explains why the derivative is c*a

6. Originally Posted by Dr.Peterson
Suppose I run twice as fast as Tim, and Tim runs 3 times as fast as Dan. Then I run 6 times as fast as Dan.

Is that the part you don't see, or is it how that relates to the derivative?
Wow.. I get it now!!!!!!!! It was that simple.. why was I so dumb. No, not how it relates to the derivative.

Thanks Dr.Peterson, and thanks Jomo!!

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