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
$$\frac {5}{ h \sqrt{3-5(x+h)} - \sqrt{3-5x}}$$

$\displaystyle f(x) = \sqrt{3-5x}$. Hence, $\displaystyle f(x+h) = \sqrt{3 - 5(x+h)}$. Therefore, we get that $$f(x+h)-f(x) = \sqrt{3 - 5(x+h)} - \sqrt{3 - 5x} = \frac{(3 - 5(x+h)) - (3 - 5x)}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}} = \frac{-5h}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}}$$
Hence, $$\frac{f(x+h)-f(x)}{h} = - \frac{5}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}}$$
Now take the limit as $h \rightarrow 0$, to get $$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = - \lim_{h \rightarrow 0} \frac{5}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}} = -\frac{5}{\lim_{h \rightarrow 0} \sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}} = - \frac{5}{2\sqrt{3 - 5x}}$$
