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

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Answer 1 · 2015-12-30T03:34:34

#f'(x)=((2x-3)(e^x+3x)-(e^x+3)(x^2-3x-6))/(e^x+3x)^2#

Quotient rule: for a function

#f(x)=(g(x))/(h(x))#

#f'(x)=(g'(x)h(x)-h'(x)g(x))/(h(x))^2#

Here:

#g(x)=x^2-3x-6#
#h(x)=e^x+3x#

#g'(x)=2x-3#
#h'(x)=e^x+3#

Thus,

#f'(x)=((2x-3)(e^x+3x)-(e^x+3)(x^2-3x-6))/(e^x+3x)^2#