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d/dx(x/1+x^2) =1-x^2/(1+x^2)^2

 Nov 16, 2014

Best Answer 

 #2
avatar+33661 
+5

I think this must be:

 

$$\frac{d}{dx}(\frac{x}{1+x^2})=\frac{1-x^2}{(1+x^2)^2}$$

 

which is correct, but the original needs brackets to make it so:

d/dx(x/(1+x2)) = (1-x2)/(1+x2)2

 Nov 16, 2014
 #1
avatar+118723 
+5

I do not know what you are asking

but

 

$$\\\frac{d}{dx}(\frac{x}{1}+x^2)\\\\
=\frac{d}{dx}(x+x^2)\\\\
=1+2x\\\\$$

 

what is the other half for?

 Nov 16, 2014
 #2
avatar+33661 
+5
Best Answer

I think this must be:

 

$$\frac{d}{dx}(\frac{x}{1+x^2})=\frac{1-x^2}{(1+x^2)^2}$$

 

which is correct, but the original needs brackets to make it so:

d/dx(x/(1+x2)) = (1-x2)/(1+x2)2

Alan Nov 16, 2014
 #3
avatar+118723 
0

Yes I should have thought of that 

 

Anon you are doing calculus - surely you can put brackets in properly !!

 Nov 16, 2014

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