Question

Chain rule/product rule confusion

Answer 1

I was doing some ODE questions like change of variable questions etc. I had a to derive the below equation. Now instantly I thought to use the product rule, but apparantly it's wrong and I had to use the chain rule. How can I work this out using the chain rule?

\[x=y ^{2}z\]

Answer 2

What is exactly the problem you are solving?

Answer 3

I do this question I needed to find the derivative of x.

Answer 4

The question does work out when the chain rule is used rather than product rule.

Answer 5

Yes, I understand, you want to include \(\large \frac{dz}{dx}\) in an equation instead of \(\large \frac{dy}{dx}\).
First, write\[y^2=\frac{x}{z}\]\[2y\frac{dy}{dx}=\frac{1}{z}-\frac{x}{z^2}\, \frac{dz}{dx}\]So you have to use \(\large \mathbf{both}\) the chain rule and the product rule to get this.

Do you understand?

Answer 6

How did you get x/z^2?

Answer 7

That's the problem. I don't see where I need to use chain rule. It seemed like a simply product rule problem.

Answer 8

Ok. \[u(x)=x,~v(x)=\frac{1}{z(x)}\]\[\frac{d}{dx}[ u(x)\, v(x) ] =u'(x)v(x)+u(x)v'(x)\]\[u'(x)=1,~v'(x)=-\frac{1}{z(x)^2}z'(x)\]\[\frac{d}{dx}[ u(x)\, v(x) ]=\frac{1}{z}-\frac{x}{z^2}\frac{dz}{dx}\]Does that make sense?

Answer 9

@Herp_Derp Yes you're just combining the derivatives at the end. I assume this is the general example, not the question? I don't have the answer with me at the moment

Answer 10

My particular question is a bit confusing.

Answer 11

Letting y^2=u and z=v

Answer 12

Now plug \(2y\frac{dy}{dx}=...\) into the equation on the left hand side, and \(y^2=...\) on the right hand side.

I don't see what your problem is...

Answer 13

@Herp_Derp The main problem was in the question itself, it caught me off guard. Any other question I would of thought of using the product rule away.