How do you differentiate #f(x)= (7 x^2 + 3 x - 6 )/ (x- 1 )# using the quotient rule?

1 Answer
Jan 31, 2016

#f'(x)=(7x^2-14x+3)/(x-1)^2#

Explanation:

The quotient rule states that for a function

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

The derivative of the function is

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

Applying this to the given function, we see that

#f'(x)=((x-1)d/dx(7x^2+3x-6)-(7x^2+3x-6)d/dx(x+1))/(x+1)^2#

Both of these derivatives can be found through the power rule.

#f'(x)=((x-1)(14x+3)-(7x^2+3x-6)(1))/(x-1)^2#

Distribute and simplify.

#f'(x)=(14x^2-11x-3-7x^2-3x+6)/(x-1)^2#

#f'(x)=(7x^2-14x+3)/(x-1)^2#