Let #f(x) = (g(x))/(h(x))#, then #f'(x) = ((g'(x) xx h(x)) - (h'(x) xx g(x)))/(h(x))^2#
Differentiating #g(x)# using the power rule:
#g'(x) = 1#
Differentiating #h(x)# using the power rule:
#h'(x) = 1#
Substituting:
#f'(x) = ((g'(x) xx h(x)) - (h'(x) xx g(x)))/(h(x))^2#
#f'(x) = ((1(x + 1)) - (1(x - 5)))/(x + 1)^2#
#f'(x) = (x + 1 - x + 5)/(x + 1)^2#
#f'(x) = (6)/(x + 1)^2#
Now, we have to plug the given #x# value into the derivative to find the slope of the tangent at that given point.
#f'(6) = 6/(6 + 1)^2#
#f'(6) = 6/49#
Now that we know the slope, let's find the value of y in #(6, y)# by plugging in #x = 6# into the original function:
#f(6) = (6 - 5)/(6 + 1)#
#f(6) = 1/7#
We currently know a point #(6, 1/7)# and the slope of the tangent, which is #6/49#.
We can now use point slope form to determine the equation of the tangent.
#y - y_1 = m(x - x_1)#
#y - 1/7 = 6/49(x - 6)#
#y - 1/7 = 6/49x - 36/49#
#y = 6/49x - 36/49 + 1/7#
#y = 6/49x - 29/49#
The equation of the tangent is #y = 6/49x - 29/49#.
Hopefully this helps!
