How do you differentiate #f(x)= ( x^2-3x-6 )/ (e^x + 2) # using the quotient rule?

Question

How do you differentiate #f(x)=  (  x^2-3x-6  )/  (e^x + 2)  # using the quotient rule?

Ezekiel Alexander · Accepted Answer

# f'(x) = ( e^x(5x+3- x^2)+4x-6) / (e^x+2)^2 # You need to use the quotient rule; # d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 # So with # f(x)=(x^2-3x-6)/(e^x+2) # we have # f'(x) = { ( (e^x+2)(d/dx(x^2-3x-6)) - (x^2-3x-6)(d/dx(e^x+2)) ) / (e^x+2)^2 } # # :. f'(x) = { ( (e^x+2)(2x-3) - (x^2-3x-6)(e^x) ) / (e^x+2)^2 } # # :. f'(x) = { ( e^x(2x-3)+2(2x-3) - (x^2-3x-6)(e^x) ) / (e^x+2)^2 } # # :. f'(x) = { ( e^x(2x-3- (x^2-3x-6))+2(2x-3) ) / (e^x+2)^2 } # # :. f'(x) = ( e^x(2x-3- x^2+3x+6)+4x-6) / (e^x+2)^2 # # :. f'(x) = ( e^x(5x+3- x^2)+4x-6) / (e^x+2)^2 #

Isabella Ackerman · Answer

undefined To differentiate $ f(x) = \frac{x^2 - 3x - 6}{e^x + 2} $ using the quotient rule, follow these steps:

1. Identify $ u(x) $ as the numerator $ x^2 - 3x - 6 $ and $ v(x) $ as the denominator $ e^x + 2 $.

2. Compute $ u'(x) $ and $ v'(x) $ separately.
   - $ u'(x) = 2x - 3 $
   - $ v'(x) = e^x $

3. Apply the quotient rule:
   $$ f'(x) = \frac{u'(x)v(x) - v'(x)u(x)}{(v(x))^2} $$

4. Substitute the values of $ u'(x) $, $ v'(x) $, $ u(x) $, and $ v(x) $ into the quotient rule formula.

5. Simplify the expression to get the derivative of $ f(x) $.