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

1 Answer
Jan 11, 2016

#(x^2 - 2x + 4 )/(x - 1 )^2#

Explanation:

for a rational function let f(x) =#u/v #

where u and v are both functions of x.Then :

#f'(x) =(v(du)/dx - u(dv)/dx)/v^2 #

differentiating using this and the chain rule gives :

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

# =( (x - 1 )(2x + 2 ) - (x^2 + 2x - 6 )(1)) /(x - 1 )^2 #

'tidying up ' the numerator :

#f'(x) (2x^2 - 2x + 2x - 2 - x^2 - 2x + 6)/(x - 1 )^2 #

#rArr f'(x) =( x^2 - 2x + 4)/(x - 1 )^2 #