Answer
$f'(4) = - \frac{1}{13}$
Work Step by Step
1. Use the Quotient Rule to Differentiate
$f(x) = \frac{(3x+1)}{(4x-3)}$
$f'(x) = \frac{[(3)(4x-3)]-[(3x+1)(4)]}{(4x-3)^{2}}$
$f'(x) = \frac{[12x-9]-[12x+4]}{(4x-3)^{2}}$
$f'(x) = \frac{12x-9-12x-4}{(4x-3)^{2}}$
$f'(x) = \frac{(-13)}{(4x-3)^{2}}$
2. Substitute the $x$ coordinate of the point into $f'(x)$
$(x, y) = (4, 1)$
$f'(4) = \frac{(-13)}{(4(4)-3)^{2}}$
$f'(4) = \frac{(-13)}{(16-3)^{2}}$
$f'(4) = \frac{(-13)}{(13)^{2}}$
$f'(4) = \frac{(-13)}{169}$
by GDC / calculator
$f'(4) = - \frac{1}{13}$
