Definition of derivative $f(x) = \sqrt{3-5x}$ I am not sure how to factor this out
$$f(x) = \sqrt{3-5x}$$
I then make it $f(x) = \frac {\sqrt{3-5(x+h)} - \sqrt{3-5x}}{h}$
I tried to multiply by the first time + the second term from the numerator which I called x and y
$$\frac {x - y}{h} \cdot \frac {x + y}{x+y}$$
which gives me
$$\frac {x^2 - y^2}{h(x+y)}$$
From here it gets very difficult
$$\frac {5}{ h \sqrt{3-5(x+h)} - \sqrt{3-5x}}$$

$$\frac{\sqrt{3-5(x+h)}-\sqrt{3-5x}}{h}\times\frac{\sqrt{3-5(x+h)}+\sqrt{3-5x}}{\sqrt{3-5(x+h)}+\sqrt{3-5x}} $$
$$=\frac{\big(3\color{Red}-5(x+\color{Red}h)\big)-\big(3-5x\big)}{h\big(\sqrt{3-5(x+h)}\color{Red}+\sqrt{3-5x}\big)}=\frac{\color{Red}-5\color{Red}h}{h\big(\sqrt{3-5(x+h)}\color{Red}+\sqrt{3-5x}\big)}.$$
After you cancel the $h$'s you can plug in $h=0$.
