Question

find f ' (2) for the function f(x) = Sqrt (3+2x) using the definition of derivative. How do I do this?

Answer 1

f(x) = (3+2x)^(1/2) ==> f'(x) = (1/2) (3+2x)^(-1/2)
f'(2) = (1/2) (3+4)^(-1/2) = 1/[2. (7)^2]

if i still remember taking derivative right :)
so hard to use with all the sqrt in the denominator

Answer 2

according to the definition\[f'(2)=\lim_{h \rightarrow 0}\frac{f(2+h)-f(2)}{h}\]

Answer 3

to use = to type :)

Answer 4

note that\[\frac{f(2+h)-f(2)}{h}=\frac{\sqrt{7+2h}-\sqrt{7}}{h}=\frac{\sqrt{7+2h}-\sqrt{7}}{h}\times\frac{\sqrt{7+2h}+\sqrt{7}}{\sqrt{7+2h}+\sqrt{7}}\]

Answer 5

do you mind finishing that one for me mukushla so i can follow all the steps you went through?

Answer 6

From where mukushla ended (rationalizing the numeration),
\[\[\frac{ (\sqrt{7+ 2h})^2-(\sqrt{7})^2 }{ h(\sqrt{7+ 2h}+\sqrt{7}) }\]
\[\frac{7+ 2h-7}{ h(\sqrt{7+ 2h}+\sqrt{7}) }\]
\[\frac{2h}{ h(\sqrt{7+ 2h}+\sqrt{7}) }\]
\[\frac{2}{ \sqrt{7+ 2h}+\sqrt{7} }\]
\[\frac{2}{ \sqrt{7}+\sqrt{7} } as h \rightarrow 0\]
\[\frac{2}{ 2\sqrt{7} } = \frac{1}{ \sqrt{7} }\]

Answer 7

thank you!