Question

Compute the derivative function f’(x) using the definition of derivative. f(x) = 3x2 + 1

Answer 1

What part are you stuck on? Just having trouble with the function notation maybe? :o
Those h's can be tough!! heh

Answer 2

the whole i need explanation on how to do it

Answer 3

Remember the formula for slope of a secant line? It's called the `Difference quotient`.\[\large \frac{f(x+h)-f(x)}{h}\]This gives us the slope of a line between two points.
What we do in calculus is, we take the "limit" of this thing.
\[\lim_{h \rightarrow 0}\large \frac{f(x+h)-f(x)}{h}\]We let the second point of our secant line get closer and closer to the first point. As h get's smaller and smaller, the line goes from being a secant line (a line through two points) to a tangent line (a line tangent to a function, in other words ~ a derivative).

All we need to do now is:
~Plug in the pieces
~Do some simplification
~Then plug 0 in for h and we'll be done.

Plugging in the pieces can be pretty difficult though. Let's do it piece by piece.

Answer 4

\[\lim_{h \rightarrow 0}\large \frac{f(x+h)-\color{orangered}{f(x)}}{h}\]Ok so we know what f(x) is, it's the thing they gave us at the start.\[\large \color{orangered}{f(x)=3x^2+1}\]

Answer 5

okay

Answer 6

yes

Answer 7

\[\lim_{h \rightarrow 0}\large \frac{\color{orangered}{f(x+h)}-f(x)}{h}\]For f(x+h), what we do is, we replace all of our x's in the problem with x+h.\[\large f(\color{royalblue}{x})=3(\color{royalblue}{x})^2+1 \qquad \rightarrow \qquad f(\color{royalblue}{x+h})=3(\color{royalblue}{x+h})^2+1\]

Answer 8

\[\large \begin{align*}\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}&=\\ \\ \\lim_{h \rightarrow 0}\frac{3(x+h)^2+1-\left(3x^2+1\right)}{h}\end{align*}\]

Answer 9

ok so it will be 6(h + x) - 3x^2 right?

Answer 10

Sorry that 2 was a typo, i fixed it on this one.

Answer 11

Should have been a power*

Answer 12

ohh ok

Answer 13

I know it's a lil confusing :c hard to get everything plugged in correctly.

Answer 14

From here, we want to expand out the square on top, and them cancel out a bunch of stuff c:

Answer 15

3(h + x)^2 - 3x^2 right?

Answer 16

the 1's cancelled? ok good c:

Answer 17

ok

Answer 18

Understand how to expand this binomial? \((x+h)^2\)

Answer 19

yes let me try it hold on

Answer 20

so i got 3h^2 + 6h x

Answer 21

After cancelling the other x's? ok good c:
See how they both have an \(h\) in them? Looks like we can go ahead and do the division with the \(h\) in the denominator.

Answer 22

yes

Answer 23

so we do this right h(3h^2 + 6h x) right?

Answer 24

\[\large \lim_{h \rightarrow 0}\frac{3h^2+6hx}{h}\]This yes? :o

Answer 25

yes

Answer 26

If you factor an h out of each term in the top, it gives you,\[\large \lim_{h \rightarrow 0}\frac{h(3h+6x)}{h}\]

Answer 27

Then you should be able to make another nice cancellation c:

Answer 28

ok

Answer 29

the h's?

Answer 30

so only 3h + 6x remain right?

Answer 31

Yah looks good :)

Answer 32

Now that we don't have any h's in the denominator, it looks like we should be able to plug in the limit value without running into any problems!!

Answer 33

ok

Answer 34

so what will be plug in? will plug in the zero right?

Answer 35

3(0) + 6x?

Answer 36

Yes good.

Answer 37

So we determined that if \(f(x)=3x^2+1\),
then \(f'(x)=6x\).

Yayyyy team \c:/

Answer 38

yes! lol thanks :)