tentukan nilai f'(1) dari f(x)= 4/(x-3) - 6/x

[tex]f(x) = \frac{4}{x - 3} - \frac{6}{x} \\ f(x) = y \\ y = \frac{4}{x - 3} - \frac{6}{x} \\ y = \frac{4x - 6(x - 3)}{(x - 3)(x)} \\ y = \frac{4x - 6}{x} \\ yx = 4x - 6 \\ yx - 4x = - 6 \\ x(y - 4) = - 6 \\ x = \frac{ - 6}{y - 4} \\ y {}^{ - 1} = \frac{ - 6}{x - 4} \\ f {}^{ - 1} (x)= \frac{ - 6}{x - 4} \\ f {}^{ - 1} (1) = \frac{ - 6}{1 - 4} \\ f {}^{ - 1} (1) = \frac{ - 6}{ - 3} \\ f {}^{ - 1} (1) = 2[/tex]