# 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

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

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$\displaystyle f(x) = \sqrt{3-5x}$. 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 $$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = - \lim_{h \rightarrow 0} \frac{5}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}} = -\frac{5}{\lim_{h \rightarrow 0} \sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}} = - \frac{5}{2\sqrt{3 - 5x}}$$
$$\frac{\sqrt{3-5(x+h)}-\sqrt{3-5x}}{h}\times\frac{\sqrt{3-5(x+h)}+\sqrt{3-5x}}{\sqrt{3-5(x+h)}+\sqrt{3-5x}}$$
$$=\frac{\big(3\color{Red}-5(x+\color{Red}h)\big)-\big(3-5x\big)}{h\big(\sqrt{3-5(x+h)}\color{Red}+\sqrt{3-5x}\big)}=\frac{\color{Red}-5\color{Red}h}{h\big(\sqrt{3-5(x+h)}\color{Red}+\sqrt{3-5x}\big)}.$$
After you cancel the $h$'s you can plug in $h=0$.