Let
#u=2x^3-1#
#v=x^2-2x#
Then
#f(x)=u/v#
Using the quotient rule, which can be found in all good google searches:
#f'(x)=(u'v-uv')/v^2#
#u'=6x#
#v'=2x-2#
#f'(x)=([6x(x^2-2x)]-[(2x^3-1)(2x-2)])/(x^2-2x)^2#
Soz, I cannot be bothered simplifying this...
#f'(x)# is the gradient function, substituting in any x value will give us the gradient of the tangent at that point.
So let's sub in x=3:
#f'(3)=([6*3(3^2-2*3)]-[(2*3^3-1)(2*3-2)])/(3^2-2*3)^2#
#rArrf'(3)=([18(9-6)]-[(54-1)(6-2)])/(9-6)^2=(18*3-53*4)/9=-158/9#
Now we want the equation of the tangent, which is a straight line of the form:
#y=mx+c#
We know #m=-158/9# but we need to solve for #c#. We need a point on the line to solve it, so sub in #x=3# into the original equation to find the y value:
#f(3)=(2*3^3-1)/(3^2-2*3)=(54-1)/(9-6)=53/3#
So the coordinates of our point on the line are (3,53/3). Sub this point into the straight line equation to solve for c:
#y=-158/9*x+c#
#rArr53/3=-158/9*3+c#
#rArr53/3=-158/3+c#
#rArrc=53/3+158/3=211/3#
So the equation of the tangent is:
#y=-158/9*x+211/3#
