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find the derivative x/x^(2) + 1 I got 2x/(x^(2)+1)
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is this \[\frac{x}{x^2+1}\] if so, you need the quotient rule
Use the quotient rule to do this question. We get u = x, v = x^2 +1 so u' = 1 and v' = 2x So the quotient rule gives \[x^2 - 1 - (2x)(x) / (x^2 + 1)^2\]. This simplifies to \[- x^2+1 / (x^2 + 1)^2\].
use \[(\frac{f}{g})'=\frac{gf'-fg'}{g^2}\] with \[f(x)=x,f'(x)=1,g(x)=x^2+1,g'(x)=2x\]
ok what did I do wrong?
Plug all in : [ (x^2 + 1) - 2x^2 ] / (x^2 + 1)^2 = ( -x^2 + 1) / (x^2 + 1)^2
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