Question

Use the limit definition of the derivative
f(x)=1/(x+3)
f'(-6)
f'(-4)

Answer 1

@zepdrix

Answer 2

please help me :(

Answer 3

\[\large \color{orangered}{f(x)=\frac{1}{x+3}}\]The Limit Definition of the Derivative:\[\large f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-\color{orangered}{f(x)}}{h}\]

See the part that I colored orange in the definition? We can plug that right in. But we also need to plug in \(f(x+h)\) correctly.

Answer 4

\[\large f(\color{royalblue}{x})=\frac{1}{\color{royalblue}{x}+3}\]\[\large f(\color{royalblue}{x+h})=\frac{1}{\color{royalblue}{x+h}+3}\]Understand that part? :)

Answer 5

sorry i will ba back in 30 mins so sorry

Answer 6

lol c: k

Answer 7

Hello are you there?

Answer 8

@zepdrix

Answer 9

\[f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\
\color{red}{f'(x)=\lim_{h\to 0}\frac{\frac{1}{x+h+3}-\frac{1}{x+3}}{h}}\\
f'(x)=\lim_{h\to 0}\frac{\frac{x+3}{(x+h+3)(x+3)}-\frac{x+h+3}{(x+3)(x+h+3)}}{h}\\
f'(x)=\lim_{h\to 0}\frac{\frac{x+3-x-h-3}{(x+h+3)(x+3)}}{h}\\
f'(x)=\lim_{h\to 0}\frac{\frac{x+3-x-h-3}{(x+h+3)(x+3)}}{h}\]
Does that help? The red part is what @zepdrix was talking about.

Answer 10

no im confused from the start, why its 1/x+h+3?

Answer 11

Like @zepdrix said,
\[f(x)=\frac{1}{x+3}, \text{ so}\\
f(x+h)=\frac{1}{(x+h)+3}\]

Answer 12

Oh i see :)

Answer 13

thank you! can i ask more question?

Answer 14

Start another thread. It's getting messy with all these replies.

Answer 15

ok, sure i will make a new state