How do you differentiate #f(x) = (x)/(x^2-x+6)# using the quotient rule?

Question

Emily Ammerman · Accepted Answer

# (6 - x^2)/(x^2 - x + 6)^2 # differentiate using the #color(blue)" quotient rule " # If f(x) =#(g(x))/(h(x))" then " f'(x) = (h(x).g'(x) - g(x).h'(x))/(h(x))^2 # #"---------------------------------------------------------------------"# g(x) = x #rArr g'(x) = 1 # and h(x) #= x^2 - x + 6 rArr h'(x) = 2x - 1 # #"------------------------------------------------------------------"# substitute these values into f'(x) #rArr f'(x) =( (x^2-x+6).1 - x(2x-1))/(x^2-x+6)^2 # #=(x^2-x+6 -2x^2+ x)/(x^2-x+6)^2 = (6-x^2)/(x^2-x+6)^2 #

Cameron Alexander · Answer

undefined To differentiate $ f(x) = \frac{x}{x^2 - x + 6} $ using the quotient rule:

1. Identify $ u(x) $ and $ v(x) $:
   $ u(x) = x $ and $ v(x) = x^2 - x + 6 $.

2. Apply the quotient rule formula:
   $ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x)\cdot u'(x) - u(x)\cdot v'(x)}{(v(x))^2} $.

3. Find the derivatives $ u'(x) $ and $ v'(x) $:
   $ u'(x) = 1 $ and $ v'(x) = 2x - 1 $.

4. Plug these into the quotient rule formula:
   $ f'(x) = \frac{(x^2 - x + 6)(1) - (x)(2x - 1)}{(x^2 - x + 6)^2} $.

5. Simplify the expression:
   $ f'(x) = \frac{x^2 - x + 6 - 2x^2 + x}{(x^2 - x + 6)^2} $,
   $ f'(x) = \frac{-x^2 + 2x + 6}{(x^2 - x + 6)^2} $.