How do you use the Quotient Rule to differentiate the function #f(x)=(x)/(x^2+1)#?

Question

Elijah Abram · Accepted Answer

I found: #f'(x)=(1-x^2)/(x^2+1)^2# The Quotient Rule tells you that if you have a function such as: #f(x)=g(x)/(h(x))# the derivative will be: #f'(x)=(g'(x)h(x)-g(x)h'(x))/[h(x)]^2# where #'# indicates derivative. In your case you get: #f'(x)=(1(x^2+1)-x(2x))/(x^2+1)^2=(x^2+1-2x^2)/(x^2+1)^2=(1-x^2)/(x^2+1)^2#

Charles Arocha · Answer

undefined To differentiate the function $ f(x) = \frac{x}{x^2 + 1} $ using the Quotient Rule:

1. Identify the numerator $ u(x) = x $ and the denominator $ v(x) = x^2 + 1 $.
2. Apply the Quotient Rule formula: 
$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$
3. Compute the derivatives:
   - $ u'(x) = 1 $ (derivative of $ x $).
   - $ v'(x) = 2x $ (derivative of $ x^2 + 1 $).
4. Plug the derivatives and the original functions into the Quotient Rule formula:
$$ f'(x) = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2} $$
5. Simplify the expression:
$$ f'(x) = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} $$
$$ f'(x) = \frac{-x^2 + 1}{(x^2 + 1)^2} $$

So, the derivative of $ f(x) = \frac{x}{x^2 + 1} $ with respect to $ x $ is $ f'(x) = \frac{-x^2 + 1}{(x^2 + 1)^2} $.