Question

Help with derivatives. Explain only, no answer.

Find the derivative of f(x) = 6x + 2 at x = 1.

Answer 1

Do you know the general formula for a tangent to a function?\[
T(x) = \frac{f(x+\Delta x) - f(x)}{\Delta x}
\]
If yes, we can get straight to business :)

Answer 2

Yes, I'm aware of the formula.

Answer 3

To be correct that's not the actual tangen, it's the slope of the function at a point \(x\)

Answer 4

But isn't it usually f(x)? I know it doesn't matter.

Answer 5

can't you just take the derivative and sub x=1 into that equation. Its the function value of the slope of the tangent line at x=1?

Answer 6

Yes

Answer 7

Wait, so I sub the line into the equation?

Answer 8

All of the other questions are the exact same thing, exept with dif numbers.

Answer 9

\[
\frac{d}{dx}f(x) = 6
\]
so it doesn't really matter at which point you look.

Answer 10

if you graphed it, there would be a straight line going across at y=6

Answer 11

so, for each x, the y=1

Answer 12

I'm not aware of this formula. Or is it something you just came up with?

Answer 13

Which formula?

Answer 14

the d/dx just means derivative or f'(x)

Answer 15

f ' (x)

Answer 16

Oh. What's the apostraphe for?

Answer 17

There are some ways to write it as @jotopia34 already mentioned

Answer 18

d/dx = derivative with respect to x. the apostrophe means f Prime at x

Answer 19

It means the first derivation with respect to x

Answer 20

that is the symbollish thingy for derivative of f(x). That's how I think of it

Answer 21

\[
\frac{d}{dx}f(x) = \frac{df}{dx} = f'(x) = D_xf
\]

Answer 22

f ' (x) = the derivative of the function f(x) in simple terms. Terms I think are easier to get than the formal definitions

Answer 23

like the captain is saying eloquently, in several, "languages" hee hee

Answer 24

Okay, thanks. CAn you guys check my work?

Answer 25

Sure

Answer 26

f ' x=8?

Answer 27

IF you replace x with 1.

Answer 28

not quite in that when you take the derivative, the derivative of a constant (number with no x) is zero. So you're adding 6+0

Answer 29

is that wrong Captain?

Answer 30

Oh, okay. So 2?

Answer 31

General derivative of a term like \(x^n\):\[
\frac{d}{dx}(x^n)= nx^{n-1}
\]

Answer 32

not quite in that when you take the derivative, the derivative of a constant (number with no x) is zero. So you're adding 6+0

Answer 33

You just said that jot. :P

Answer 34

You are right with the 6 part, only you don't add 2 because the derivative of 2 = 0

Answer 35

So if you take the derivative of a constant you could look at that as \(ax^0\) now try it and plug that into the fomula

Answer 36

Oh, that's what you meant.

Answer 37

yes, if you're referring to my 6+0 comment

Answer 38

So any constant in the original equation becomes 0?

Answer 39

yes

Answer 40

\[
\frac{d}{dx}(ax^0) = 0 a x^{-1} = 0
\]

Answer 41

I am btw.

Answer 42

Okay. Thanks!

Answer 43

I had trouble with this too at the beginning. Formal definitions are tricky to understand.

Answer 44

I think of the constant thing as there is no x in back of it, so it is zero

Answer 45

So if it was f(x)=-9/x with x=6, it would be -9/6? (I would simplify)

Answer 46

-3/2

Answer 47

if the x i in the denominator it's different :)

Answer 48

So may rules!

Answer 49

:(

Answer 50

many*

Answer 51

the x in the denominator is like saying x^-1\[x^-1\]

Answer 52

for example, instead of 2, its (-1). You do the same derivative as a positive number but don't forget to =bring the negative sign with you

Answer 53

\[
\frac{d}{dx}\left( \frac{1}{x}\right)=\frac{d}{dx}\left( x^{-1}\right) = -1x^{-2} = \frac{-1}{x^2}
\]

Answer 54

You have to do do prodt rule with x^-1 or the quotient rule. \[x^-1 = -1(x)^{-1-1}\]\[=-x ^{-2}\]

Answer 55

which is the same as -1/x^2\[\frac{ -1 }{ x^2 }\]

Answer 56

does any of this make sense?

Answer 57

Not really.

Answer 58

okay, easy to do. part by part

Answer 59

You start with f(x)=-9/x right?

Answer 60

so next, you want to take the derivative. There are 2 ways of doing it. Do you know the product rule or the quotient rule?

Answer 61

Quotient rule.

Answer 62

Great. So that rule is: (Derivative of the top x bottom), minus (the derivative of the bottom x the top) all over the (bottom part)^2 squared

Answer 63

Wait, I don't get it.

Answer 64

We have in the quotient a top/bottom. Numerator function / denominator function

Answer 65

numerator =-9
denom. = x

Answer 66

numerator = top function
denom = bottom function

Answer 67

so when I say derivative of the top, I mean derivative of whatever function is in the numerator. In your case it is -9

Answer 68

or whatever function is on "top"

Answer 69

Oh, okay.

Answer 70

so its
(deriv. top function)(bottom function with no derivative taken) - (deriv. of bottom function)(top function with no derivative taken)

Answer 71

There is a general formula, in which you can plug it in:\[
\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g-g'f}{g^2}
\]
Take \(f(x) = \frac{9}{x}\) and plug it in:\[
\frac{d}{dx}\left(\frac{9}{x}\right) = \frac{0x-9}{x^2} = \frac{-9}{x^2}
\]

Answer 72

the math notation looks sometimes like top=f'(x) (g(x) - g'(x) * f(x) / g(x)^2\[\frac{ 0(x) - (1)(-9) }{ x^2}\]
\[=\frac{ -9 }{ x^2 }\]

Answer 73

in above, my -9 should just be 9. sorry!

Answer 74

Now I'm very confused.

Answer 75

i am sorry

Answer 76

I think i will leave jotopia to it as it can get confusing when 2 people are talking bout the same topic. If u need anything simply message me :) And good luck with derivatives.

Answer 77

what works for me, is if you ask a step by step question, or say if something doesn't make sense in the steps.

Answer 78

for example, what's the first thing you are going to do to take this derivative

Answer 79

I'm not sure.

Answer 80

okay, first is to figure out, what "formula" you need to take the derivative
Do you know what that is?

Answer 81

No.

Answer 82

Okay, because we have a fraction, we need to us the formula that is for a function that is any type of fraction. Do u know which that is?

Answer 83

No, I am completely new with derivatives with fractions.

Answer 84

Very slow comp.

Answer 85

okay. To do derivative of a fraction, there is a "formula". The "formula" IS the quotient rule

Answer 86

Here is the formula:

Answer 87

dang, can't see it. ok that reads:
\[f(x) =\frac{ 9 }{ x } \]

this is your question

Answer 88

\[\frac{ d }{ dx } f(x) = f'(x)\]
What the above is saying is
the derivative (d/dx) of your function , f(x) is the same notation as f'(x)

Answer 89

All that is saying is that now we are going to take the derivative of your function 9/x

Answer 90

for the picture, the quotient rule is
\[\frac{ d }{ dx }\left[ \frac{ f(x) }{ g(x) } \right]=\frac{( f'x)(gx)-(g'x)(fx) }{ (gx)^2 }\]

Answer 91

quiz time. What is the f(x) of your question

Answer 92

9/x

Answer 93

actually, when you have a fraction, you have to separate it into 2 parts. top part and a bottom part. Each part gets a function name.

Answer 94

So looking at the picture of the quotient rule equation, which is on the top, f(x) or g(x)

Answer 95

?

Answer 96

Does it nmake any sense that we have to divide the fraction into 2 parts

Answer 97

Yes.