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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

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

2 Answers 2

4

$\displaystyle f(x) = \sqrt{3-5x}$. 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 $f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = - \lim_{h \rightarrow 0} \frac{5}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}} = -\frac{5}{\lim_{h \rightarrow 0} \sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}} = - \frac{5}{2\sqrt{3 - 5x}}$

7

$\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 $h=0$.