1. ## Learning Derivatives

Beginner. Working on this problem: Use the definition of a derivative function to calculate f'(4) where f(x) = x + 3.
Have made attempt using lim h->0 [f(x+h) - f(x)]/h but unsure if answer is correct. Need explanation. Help!

2. Originally Posted by Yami Sorceress
Beginner. Working on this problem: Use the definition of a derivative function to calculate f'(4) where f(x) = x + 3.
Have made attempt using lim h->0 [f(x+h) - f(x)]/h but unsure if answer is correct. Need explanation. Help!
$
\lim_{h\to0}\frac{[(x+h)+3]-(x+3)}{h}$

Now simplify and take the limit? Got it?

Hint: the limit is constant

3. There is an equivalent definition that is sometimes useful:

$f'(c) = \lim_{x \to c} \frac {f(x) - f(c)}{x-c}$

so $f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x-4}$ $= \lim_{x \to 4} \frac {(x+3) -7}{x-4}$ $= \lim_{x \to 4}\frac {x-4}{x-4} = 1$

4. ## then ...

OK, so for clarity I could write:

lim h->0 (x+3)+h - (x+3) / h

simplify:
lim h->0 h/h = 1 ?

Where does the f ' (4) fit in?

5. Originally Posted by Yami Sorceress
OK, so for clarity I could write:

lim h->0 (x+3)+h - (x+3) / h

simplify:
lim h->0 h/h = 1 ?

Where does the f ' (4) fit in?
$f'(4) = \lim_{h \to 0} \frac {f(4+h) - f(4)}{h}$

6. Thank you for the clear example.

7. Originally Posted by Yami Sorceress
Where does the f ' (4) fit in?
What you must understand is that $f'(x)=1$ for ALL x. So, $f'(4)=1$, $f'(132,345,225)=1$. All x. This makes sense right? because the slope of a line is constant right?