dy/dx = dy/du * du/dx

burgerandcheese

burgerandcheese

Junior Member
Joined
Jul 2, 2018
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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 :confused:
 
burgerandcheese said:
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 :confused:
Click to expand...
y = c(ax + b) + d = cax+cb+d. So dy/dx =ca
 
burgerandcheese said:
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 :confused:
Click to expand...

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?
 
Dr.Peterson said:
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?
Click to expand...

Wow.. I get it now!!!!!!!! It was that simple.. why was I so dumb.:oops: No, not how it relates to the derivative.

Thanks Dr.Peterson, and thanks Jomo!! :D :D
 
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