# Derivatives Quotient Rule

• May 17th 2010, 12:47 PM
oregon88
Derivatives 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 PM
e^(i*pi)
Quote:

Originally Posted by oregon88
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.

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

Wolfram gets the same answer I do. If you are unsure about how to use the quotient rule here the chain rule works equally well

Spoiler:
By the chain Rule:

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

$f'(x) = (x^2+1)^{-2} \cdot 2x = \frac{2x}{(x^2+1)^2}$
• May 17th 2010, 01:09 PM
oregon88
$f(x)=-x/x^2+1$

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

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

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

Is what I came up with
• May 17th 2010, 01:20 PM
e^(i*pi)
Quote:

Originally Posted by oregon88
$f(x)=-x/x^2+1$

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

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

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

Is what I came up with

Hold on, you said $f(x) = -\frac{1}{x^2+1}$ 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

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

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

$f'(x) = \frac{-x^2-1 + 2x^2}{(x^2+1)^2} = \frac{x^2-1}{(x^2+1)^2}$
• May 17th 2010, 01:34 PM
oregon88
The original problem is

$
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)