The definition of derivative is
f'(x) = \lim_{x\to0}\frac{f(x+h)-f(x)}{h}
so, if f(x) = \frac{5-x}{x}, the definition leads to
f'(x) = \lim_{x\to0}\frac{\frac{5-x-h}{x+h}-\frac{5-x}{x}}{h}
Manipulate the numerator to get
\frac{5-x-h}{x+h}-\frac{5-x}{x} = \frac{(5-x-h)x - (5-x)(x+h)}{x(x+h)
=\frac{5x-x^2-hx - 5x+x^2-5h+hx}{x(x+h)} = \frac{-5h}{x(x+h)}
so,
f'(x) = \lim_{x\to0}\frac{\frac{5-x-h}{x+h}-\frac{5-x}{x}}{h} = \frac{-5h}{hx(x+h)} = \frac{-5}{x(x+h)}
When h \to 0, x+h \to x, so
f'(x) = \lim_{x\to0}\frac{-5}{x(x+h)} = \frac{-5}{x^2}
