Question

How the heck do I figure out the equations for the first derivative.. Simple explanation please because I am extremely confused

Answer 1

Do you mean stuff like
\[f(x) = x^n\]\[f'(x) = nx^{n-1}\]

Answer 2

I don't even know what I mean I'm just lost

Answer 3

The derivative gives the rate of change. Or in other words the slope.

Answer 4

Oh.. I just don't get how it relates to the relative extrema like f(x)=3x^2+5x+2.. Like when you end up with 3 equations or something

Answer 5

Remember \(y = mx + b\)? M is the slope.
So let's write that as \(f(x) = mx^1 + b\) and take the derivative using that formula above. That's \(f'(x) = 1mx^{1-1} = m\) so, yeah, the slope. (B just goes away because it's a constant)

Answer 6

That is just where the derivative changes sign. In other words the slope stops going up and starts going down (or stops going down and starts going up)

Answer 7

So what would that look like with the equation I wrote? And what would the relative max be?

Answer 8

Well first we have to take the derivative of \(f(x) = 3x^2 + 5x +2\).
We can do that with the formula above! So do that first.

Answer 9

That's what I don't know how to do lol

Answer 10

I just showed you. You subtract 1 from the exponent and multiply it by the coefficient.

Answer 11

Ohhhh!

Answer 12

The derivative of \(2x^3\) is \(6x^2\) for example because it's \((2 \times 3)x^{3-1}\)

Answer 13

So you just do that for every number then?

Answer 14

Right. And get rid of anything with no x.

Answer 15

(Or to be fancy about it, if \(h(x) = f(x) + g(x)\) then \(h'(x) = f'(x) + g'(x)\) )

Answer 16

Okay so hold on..

Answer 17

What if it's 5x? Would that stay as 5x?

Answer 18

No, \(5x\) means \(5x^1\)

Answer 19

So what's \((1 \times 5)x^{0}\) ?

Answer 20

Isn't that the same thing? & so far I have (-3 x 2)(x^1) is that right?

Answer 21

& would that be 5x?

Answer 22

Was it negative? It was positive before. If so, so far so good.

Answer 23

Yes it was negative.. Typo sorry

Answer 24

And, no. \(x^0 = 1\). The derivative of \(5x\) is just 5. Like I showed above, the derivative of y=mx + b is just m.

Answer 25

Ohhhhhh that makes sense I forgot anything to the power of 0 is 1.. So I have (-3 x 2)(x^1)+5x^1?

Answer 26

You forgot to get rid of the x again. :P

Answer 27

Wait.. What? :p

Answer 28

It's just \(6x + 5\)

Answer 29

Because it's \(6x^1 + 5x^0\)

Answer 30

-6, rather.. I forgot the negative

Answer 31

Ohhhhh wait! It's 5x^0 because its 5x^1-1 right!?

Answer 32

Yes.

Answer 33

Multiplying times 1 and thinking of it as \(x^0\) and stuff is tedious and as you learn it you'll just drop the x but while learning it helps you understand the process

Answer 34

Okayy that makes sense.. So would I write it as f(x)=-6x+5 or f'(x)=-6x+5 and what does the ' mean anyways?

Answer 35

The ' is read "prime" and it just means "this is a derivative."

Answer 36

And also.. How do you figure out if its a critical number? Like what does that have to do with anything?

Answer 37

And ohhh

Answer 38

Well, now we know the slope!

Answer 39

Okay! So what now?

Answer 40

Ok, well, imagine the graph of \(-3x^2 + 5x + 2\).. It's going to be an upside down parabola.

Answer 41

Okay

Answer 42

So, where on the parabola is the largest that function is ever going to get? (Not the exact value, just, where on the shape is it)

Answer 43

The tip of the curve?

Answer 44

Right! And what's the slope there? (imagine a tangent line)

Answer 45

-6?

Answer 46

No. The curve flattens out!

Answer 47

So what does that mean?

Answer 48

It means the slope is 0

Answer 49

But how do you know that

Answer 50

How do I know what?

Answer 51

That it's 0.. You lost me :p

Answer 52

Here is my really bad not to scale parabola

Answer 53

Okay

Answer 54

The slope at any given point is the same as the slope of the tangent line

Answer 55

You're gunna hate me... What's a tangent line?:p

Answer 56

It gets a little circular... but the tangent line is the line you draw next to the curve that intersects that point and has the same slope as the curve at that point.

Answer 57

Would it help to try a completely different way of thinking about it?

Answer 58

Probably :p

Answer 59

Ok. You see how the parabola goes up and then goes down? In other words, the slope is positive, then it's negative.

Answer 60

Yes

Answer 61

Well, since it's a continuous line, to go from positive to negative means it changed direction. So it had to have been 0 at some point.

Answer 62

And whenever that happens, it WILL be 0?

Answer 63

For the most part. For polynomial functions with smooth curves like this, definitely.

Answer 64

Okay so that means its a critical number?

Answer 65

Right

Answer 66

& Why is that important

Answer 67

You're trying to find the relative max, right?