Question

Product rule question.

Answer 1

I got this ODE problem and that the product rule had to used. I got the solution from teacher and I have no idea where he gets the dy/dx from

Answer 2

I know how to use the product rule but no idea where my teachers gets dy/dx

Answer 3

chain rule

Answer 4

Chain rule on the entire of LHS?

Answer 5

on each term

Answer 6

I'm confused, from looking at it, it just looks like product rule. Unless I am missing something obvious.

Answer 7

two products multiplying by each other.

Answer 8

It is the product rule, but for each individual derivative he is using the chain rule.
What is d/dx of y^2?
2 y dy/dx

Answer 9

We have y^2 times z
The derivative is z times d/dx of y^2 + y^2 times d/dx of z
on the rhs, d/dx of x = 1

Answer 10

I would of automatically said 2y

Answer 11

Do you understand now? Read both my previous posts.

Answer 12

I don't understand your second post.

Answer 13

...if you d/dx \(y^2 z\) wont it just be 0? o.O am i missing something here?

Answer 14

@lgbasallote both y and z are functions of x

Answer 15

ohh...should've said that sooner

Answer 16

anyway...what ironictoaster did looks right to me...where is it wrong @telliott99 ?

Answer 17

We have y^2z, both y and z functions of x
We use the product rule
y^2 times dz/dx + z times d/dx(y^2)
d/dx(y^2) = 2 y dy/dx by the chain rule

Answer 18

I am trying get a step by step of the chain rule part to make sure I know what I am doing in future, I just keep getting 2y on Wolfram..

Answer 19

I mean it makes sense to me, but I never would of copped on to this.

Answer 20

Let u = y^2
We want d/dx ( uz )
The term we are worrying about is
z times du/dx
du/dy = 2y
du/dx = du/dy dy/dx

Answer 21

The last line is the definition of the chan rule right?

Answer 22

No wait it's not!