How do you differentiate #f(x) = (5x)/(3x^2-4x+6)# using the quotient rule?

1 Answer
mason m
Jan 20, 2016

#f'(x)=(-15x^2+30)/(3x^2-4x+6)^2#

Explanation:

For a function #f(x)=(g(x))/(h(x))#, the quotient rule states that the derivative of the function is

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

In this scenario, we know the following:

#g(x)=5xcolor(white)(xxxx)=>color(white)(xxxx)g'(x)=5#

#h(x)=3x^2-4x+6color(white)(xxxx)=>color(white)(xxxx)h'(x)=6x-4#

Thus, plugging these into the quotient rule formula,

#f'(x)=(5(3x^2-4x+6)-5x(6x-4))/(3x^2-4x+6)^2#

Simplify.

#f'(x)=(15x^2-20x+30-30x^2+20x)/(3x^2-4x+6)^2#

#f'(x)=(-15x^2+30)/(3x^2-4x+6)^2#

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