What is the Chain Rule and How Do You Apply It?

[schlynn]

Homework Statement

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

Homework Equations

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]

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

<br /> \frac{dy^2}{dx}=\frac{dy^2}{dy}\frac{dy}{dx}<br />

Can you work this out?

 

[schlynn]

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]

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?

 

[schlynn]

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.

 

[Cyosis]

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

 

[schlynn]

Yeah...its 2x.

 

[Cyosis]

No it's 2 x \exp x^2. Either way if you have a composite function, f(g(x)) then (f(g(x)))&#039;=\frac{df(g(x))}{dg(x)}\frac{dg(x)}{dx}. Have you ever seen this?

 

[schlynn]

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]

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.

 

[schlynn]

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

 

[Cyosis]

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

 

[schlynn]

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]

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.

 

[schlynn]

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]

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.

 

[schlynn]

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

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

 

[Cyosis]

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)&#039;=(u^2)&#039;*u&#039;=2u*u&#039;=2y(x)y&#039;(x)=2y*y&#039;.

 

[Mark44]

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?

 

[Dhaos]

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

 

[schlynn]

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.

 

[Cyosis]

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.

 

[schlynn]

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?

 

[Cyosis]

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

 

[schlynn]

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

 

[Mark44]

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 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?

 

[Bohrok]

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.

 

[schlynn]

Just checked the thread now. Is the answer...

1/2 * 1/cos²x? Or the same with just square root of x?

 

[Cyosis]

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!