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

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Answer 1 · 2016-11-01T21:00:27

# 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 #