In the problem f(x) = -1/x^2+1 finding the derivative using the quotient rule.

the solution would be -x^2+2x-1/(x^2+1)^2?

It seems right but im not sure.

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- May 17th 2010, 12:47 PMoregon88Derivatives Quotient Rule
In the problem f(x) = -1/x^2+1 finding the derivative using the quotient rule.

the solution would be -x^2+2x-1/(x^2+1)^2?

It seems right but im not sure. - May 17th 2010, 12:58 PMe^(i*pi)
- May 17th 2010, 01:09 PMoregon88
$\displaystyle f(x)=-x/x^2+1$

$\displaystyle f'(x)=(x^2+1)d/dx(-x)-(-x)d/dx(x^2+1)/(x^2+1)^2$

$\displaystyle (x^2+1)(-1)-(-x)(2x)/(x^2+1)^2$

$\displaystyle -x^2+2x-1/(x^2+1)^2$

Is what I came up with - May 17th 2010, 01:20 PMe^(i*pi)
Hold on, you said $\displaystyle f(x) = -\frac{1}{x^2+1}$ in the OP. Which one is correct?

Your differentiating is correct but your simplification on the third line is not - there should be no x term. Plus be more clear with your syntax using brackets

$\displaystyle u = -x \: \rightarrow \: u' = -1$

$\displaystyle v = x^2+1 \: \rightarrow \: v' = 2x$

$\displaystyle f'(x) = \frac{-x^2-1 + 2x^2}{(x^2+1)^2} = \frac{x^2-1}{(x^2+1)^2}$ - May 17th 2010, 01:34 PMoregon88
The original problem is

$\displaystyle

f(x) = \frac {-x}{x^2+1}

$

Sorry about that, I see where I messed up on the third line I did not factor the (-x)(2x) on my paper. Thanks for the help e^(i*pi)