Derivatives Quotient Rule

May 2010
25
0
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.
 

e^(i*pi)

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Feb 2009
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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.
I don't get that answer, perhaps you could post your working?

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

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

By the chain Rule:

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

\(\displaystyle f'(x) = (x^2+1)^{-2} \cdot 2x = \frac{2x}{(x^2+1)^2}\)
 
Last edited:
May 2010
25
0
\(\displaystyle f(x)=-x/x^2+1\)

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

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

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

Is what I came up with
 

e^(i*pi)

MHF Hall of Honor
Feb 2009
3,053
1,333
West Midlands, England
\(\displaystyle f(x)=-x/x^2+1\)

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

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

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

Is what I came up with
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

\(\displaystyle u = -x \: \rightarrow \: u' = -1\)

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


\(\displaystyle f'(x) = \frac{-x^2-1 + 2x^2}{(x^2+1)^2} = \frac{x^2-1}{(x^2+1)^2}\)
 
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May 2010
25
0
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)