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

## Homework Statement

x2+y2=1
I want to differentiate this equation. I know that the answer is 2x+2y*y'=0.

The chain rule.

## The Attempt at a Solution

I don't understand how you get 2y*y' from y2. Shouldn't it just be 2x+2y=0?

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

$$\frac{dy^2}{dx}=\frac{dy^2}{dy}\frac{dy}{dx}$$

Can you work this out?

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I think that I see it now. Is it....

2y=2y*y'*1?

That is pretty much a random shot in the dark, but I see how its 2y*y' because its dy2/dy.

Cyosis
Homework Helper
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 $\frac{dy}{dx}$?

but I see how its 2y*y' because its dy2/dy

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?

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I don't understand it, but thats what I was asking origianaly. Can you explain it since I don't get it? Understood everything up untill this.

Cyosis
Homework Helper
Okay, to check your knowledge of the chain rule do you know how to differentiate, lets say $\exp{x^2}$?

Yeah...........its 2x.

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

Yes, I have seen this. The composition Rule. I thought that exp meant example. What function is it. And for example sin(t)10=9*cos(t) right?

Cyosis
Homework Helper
Hehe, no exp means the exponent e. The derivative of $\sin^{10}(t)=10\sin^9(t) \cos(t)$.

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.

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Something like put y2 in for x2 so its (y2)2 so its 2(y2), then what?

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

I'm lost. I have no idea what to do. I am trying but where does the other functions come from? I usually get everything really easily but then when I run into something like this i'm wondering around in the dark. Can you explain how the numbers work? Like how it all works out and why.

Mark44
Mentor
Yes, I have seen this. The composition Rule. I thought that exp meant example. What function is it. And for example sin(t)10=9*cos(t) right?

What Cyosis wrote in post 8 is the Chain Rule, using Newton's notation. There is no composition rule, althought the Chain Rule is used for finding the derivative of a function composition. In post 2 Cyosis used a different form of the chain rule, that uses Leibniz notiation.

But I still don't get it. This isn't a homework question. You guy can just explain it to me. I need to learn this to be able to continue with my study of calculus. I have tried but I fail to see where the composition part comes in. All I have is y2.

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

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

But I still don't get it. This isn't a homework question. You guy can just explain it to me. I need to learn this to be able to continue with my study of calculus. I have tried but I fail to see where the composition part comes in. All I have is y2.

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

Havn't learned the derivative of ln but pretty sure its e right? Because e is the inverse of ln. Like ln(ex)=x. Is is 2e?

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I reliase that that is wrong now. Because e is the derivative and anti-derivative of e.

Cyosis
Homework Helper
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 $y(x)^2$. Make the substitution u=y(x), then using the chain rule $(y(x)^2)'=(u^2)'*u'=2u*u'=2y(x)y'(x)=2y*y'$.

Mark44
Mentor
But I still don't get it. This isn't a homework question. You guy can just explain it to me. I need to learn this to be able to continue with my study of calculus. I have tried but I fail to see where the composition part comes in. All I have is y2.
But the trouble is, y is a function of x, and you want the derivative with respect to x. If you were differentiating with respect to y, it would be easy -- d/dy(y2) = 2y and you're done.

But what you want is d/dx(y2) which is not 2y.

To use a better example, let y = (cos(x))3. To get a y value, you have to start with an x value, then evaluate the cosine of that x value, then cube that value. When you take the derivative with respect to x, the chain rule takes the composition into account.

So dy/dx = d( (cos(x))3)/d( cos(x)) * d(cos(x))/dx

In words, the derivative of y with respect to x is the derivative of (cos(x))3 with respect to cos(x) times the derivative of cos(x) with respect to x.

So dy/dx = 3 (cos(x))2 * (-sin(x))

Is that any clearer?

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

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The equation for the chain rule is (un)'*u'? Why didn't you say that? It makes since now. The dy/dx notation is confusing. I know that it's y with respect to x but it's not as easy to visualize as u'.

If I am still mistaken I still don't get it. Can you explain it with out y(x) and dy/dx? Just use y' and y2. The video on that site helped but I already knew that. Doing this with trig functions is super easy.

Cyosis
Homework Helper
Why didn't I say that? First of all it is the worst form of the chain rule, it's not very clear. For example the first accent means differentiation with respect to u and the second one with respect to x. Secondly I asked you multiple times how you used the chain rule yourself.

Well not that I see how it works. Can someone help me explain the underlying math? I want to understand HOW. Not just a equation. I know that you guys have been trying to do that all along but maybe a different approach?

Thank you all for being patient with me. Now I can learn more.

Mark44
Mentor
The equation for the chain rule is (un)'*u'? Why didn't you say that? It makes since now. The dy/dx notation is confusing. I know that it's y with respect to x but it's not as easy to visualize as u'.
It's not "y with respect to x"; it's the derivative of y with respect to x. Once you understand what the symbols mean, the dy/dx notation of Leibniz is much superior, IMO, to the Newton-style notation. For one thing dy/dx is more suggestive of the difference quotient (f(x+h) - f(x))/h or $\Delta y/\Delta x$. For another, it tells you exactly what the independent variable is; IOW, what the variable with respect to which you're differentiating. As Cyosis pointed out, that information isn't present in u'.
If I am still mistaken I still don't get it. Can you explain it with out y(x) and dy/dx? Just use y' and y2. The video on that site helped but I already knew that. Doing this with trig functions is super easy.

Since you're up to speed with trig functions, if y = tan($\sqrt{x})$, can you find dy/dx?

FWIW, You should know that
$$\frac {d} {dx} y = \frac{dy} {dx}$$

Then
$$\frac {d} {dx} y^2 = \frac {d} {dx} (y*y)$$
and use the product rule.

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No this is not right. First we differentiate $\tan \sqrt{x}$ with respect to $\sqrt{x}$ if you will which yields $sec^2 \sqrt{x}$. Then we differentiate $\sqrt{x}$ with respect to x which yields $\frac{1}{2\sqrt{x}}$.
Answer: $$\frac{1}{2\sqrt{x} \cos^2 \sqrt{x}}$$