Chain rule, y^2'=2y*y'?

You're differentiating with respect to x right? Now lets apply the chainrule.

[tex]
\frac{dy^2}{dx}=\frac{dy^2}{dy}\frac{dy}{dx}
[/tex]

Can you work this out?

 

Sorry but how can you think you see it now when it's a random shot in the dark. It seems you don't really understand how the chain rule works. What is [itex]\frac{dy}{dx}[/itex]?

No dy^2/dy=2y.

Do you understand the form I have written the chain rule in in my previous post. If not how do you use the chain rule usually?

 

Okay, to check your knowledge of the chain rule do you know how to differentiate, lets say [itex]\exp{x^2}[/itex]?

 

No it's [itex]2 x \exp x^2[/itex]. Either way if you have a composite function, f(g(x)) then [tex](f(g(x)))'=\frac{df(g(x))}{dg(x)}\frac{dg(x)}{dx}[/tex]. Have you ever seen this?

 

Hehe, no exp means the exponent e. The derivative of [itex]\sin^{10}(t)=10\sin^9(t) \cos(t)[/itex].

You can view y as a composite function in away as well since we are differentiating y with respect to x, y must be a function of x. Therefore we want to find the derivative of y(x)^2. Try to calculate the derivative now.

 

No apply the chain rule I have listed in post 8.

 

I am trying to find a form of the chain rule he recognizes.

Schlynn could you differentiate [itex]\ln x^2[/itex] with a method you're used to? Please show all the steps.

You differentiate y^2 with respect to x, not y! That's where the chain rule comes into play, it really is [itex]y(x)^2[/itex]. As for just explaining it, I am trying but just giving you the result won't do you any good.

 

If you haven't learned the derivative of the logarithm ignore my previous post.

Lets try it again another way!

We want to find the derivative of [itex]y(x)^2[/itex]. Make the substitution u=y(x), then using the chain rule [itex](y(x)^2)'=(u^2)'*u'=2u*u'=2y(x)y'(x)=2y*y'[/itex].

 

I think this can help you http://www.calculus-help.com/funstuff/phobe.html" [Broken], I recommend you to watch the whole set of explanations, they are very clear :P