# Derivatives Quotient Rule

#### 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.

#### e^(i*pi)

MHF Hall of Honor
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.

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

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:

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

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

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#### oregon88

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

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

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

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

Is what I came up with

#### e^(i*pi)

MHF Hall of Honor
$$\displaystyle f(x)=-x/x^2+1$$

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

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

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

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

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

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

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

oregon88

#### oregon88

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)