Mastering the Chain Rule: A Quick Guide for Calculus Students

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Homework Statement


y = 2x / (1+x^2)^2

Find dy/dx

Homework Equations


Chain rule


The Attempt at a Solution


I completely forgot how to apply the chain rule.. I mean, I can always apply the quotient rule, but I'm sure this is 1000 times easier if you can apply the chain rule. Do you do something like

u = 1+x^2
du = 2x

so

y = du/u

But I may be getting confused with substitution rule with integration.. it's been a while since I touched calculus.. any suggestions?
 
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Chain rule goes something like
dy/dx=dy/du*du/dv*...*df/dx .

It is usually used when you when you have a nested combination of functions, ie functions within functions.

For your question, you need to use both the quotient rule as well as the chain rule ( (1+x^2)^2, which is the funtion 1+x^2 within a squaring function ) .

Can you finish your problem now ?
 
arunbg said:
Chain rule goes something like
dy/dx=dy/du*du/dv*...*df/dx .

It is usually used when you when you have a nested combination of functions, ie functions within functions.

For your question, you need to use both the quotient rule as well as the chain rule ( (1+x^2)^2, which is the funtion 1+x^2 within a squaring function ) .

Can you finish your problem now ?

Right.. that's the BRUTE force way to do it.

I was wondering if there was a way to finish this without even applying the quotient rule
 
if you don't want to use the quotient rule you can bring the denominator up top so:y= 2x (1+x^2)^{-2} now use the product rule
 
l46kok said:
Right.. that's the BRUTE force way to do it.

I was wondering if there was a way to finish this without even applying the quotient rule

Yes, just write the expression as 2x(1+x2)-2, and use the product rule on this.
 
suspenc3 said:
if you don't want to use the quotient rule you can bring the denominator up top so:y= 2x (1+x^2)^{-2} now use the product rule

I guess that's true, but when I saw the relating terms, I was thinking this could be solved by ONLY using chain rule.

I guess it's impossible.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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