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

• schlynn
In summary, the conversation revolved around differentiating the equation x^2+y^2=1 using the chain rule. The formula for the chain rule was discussed and applied to the given equation. The concept of a composite function was also explained, with the substitution u=y(x) used to simplify the process. Additional resources were also recommended for better understanding of the chain rule.

## 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?

You're differentiating with respect to x right? Now let's 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.

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 that's what I was asking origianaly. Can you explain it since I don't get it? Understood everything up until this.

Okay, to check your knowledge of the chain rule do you know how to differentiate, let's say $\exp{x^2}$?

Yeah...its 2x.

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?

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?

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.

schlynn said:
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.

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.

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 $(y(x)^2)'=(u^2)'*u'=2u*u'=2y(x)y'(x)=2y*y'$.

schlynn said:
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 without 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.

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?

You can find a proof for the chain rule here http://math.rice.edu/~cjd/chainrule.pdf.

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

schlynn said:
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'.
schlynn said:
If I am still mistaken I still don't get it. Can you explain it without 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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1/2 * 1/cos2x? Or the same with just square root of x?

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: $$\frac{1}{2\sqrt{x} \cos^2 \sqrt{x}}$$

You should really review the chain rule thoroughly!

## What is the chain rule in calculus?

The chain rule is a mathematical rule used in calculus to find the derivative of a composite function. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

## How does the chain rule apply to y^2' = 2y*y'?

In this equation, y^2 is the outer function and 2y is the inner function. Applying the chain rule, we get y^2' = 2y * y', where y' represents the derivative of y. This equation can be simplified to y' = 2y, which is the derivative of the original equation.

## Can the chain rule be used to find the derivative of any composite function?

Yes, the chain rule can be used to find the derivative of any composite function. It is a fundamental rule in calculus and is used extensively in solving various mathematical problems.

## What are some real-world applications of the chain rule?

The chain rule has many applications in physics, engineering, and economics. For example, it can be used to find the rate of change of a quantity over time, such as the speed of an object moving along a curved path. It is also used in optimization problems to find the maximum or minimum value of a function.

## How can I practice and improve my understanding of the chain rule?

To improve your understanding of the chain rule, you can practice solving problems that involve composite functions and their derivatives. You can also use online resources, such as tutorials and practice exercises, to further enhance your knowledge and skills in applying the chain rule.