Question

Calculus:

What does it mean to take the derivative of something with respect to something?

I know how I can compute the derivative of something, but what does it mean to take it with respect to something else?

Answer 1

It means solve for the particular derivative

Answer 2

how does this change when we change this other thing

Answer 3

can you be more concrete?

Answer 4

how fast is the height of a ball changing after 5 seconds of being dropped from 300 meters?

Answer 5

taking the derivative with respect to x means finding the change in the x derivative or partial derivative

Answer 6

a derivative is a rate of change; the "respect to" is relating it to some variable

Answer 7

so like finding the equation for the tangent of a function at any point?

Answer 8

slope of tangent*

Answer 9

how long can you last in a boxing ring ... with respect to Mike Tyson?

how long can you last in a boxing ring ... with respect to Gary Bussey?

Answer 10

so you wouldn't solve for the equation but for the particular value

Answer 11

the derivative gives you another equation, which can be used to determine the slope of the tangent (the measure of the rate of change) at a given value

Answer 12

I know that

Answer 13

But when you're taking it with respect to something, you're basically asking what the slope of that tangent is at that point, instead of the equation of the slope of the tangent at any point?

Answer 14

correct

Answer 15

ok thanks!

Answer 16

yw

Answer 17

Hmm, let me ask another question, because I think I still might be understanding it wrongly:

derivative of ln(x^2) with respect to x^2, according to khan academy, is 1/(x^2)

But if you solve for the particular value, wouldn't that actually mean you have to find the derivative of ln( (x^2)^2 )?

Answer 18

if we let: a = x^2

ln(a)' = 1/a

Answer 19

if we want to know what it is wrt "x"; then let a(x) = x^2

ln(a)' = a'/a

Answer 20

Well I see I'm interpreting it wrong, I actually have to find the derivative of ln(x^2), but then inside the derivative, substitute x^2 for x

Answer 21

yet, derivative of ln(x^2) is actually (1/(x^2))*2x

Answer 22

keep in mind that the chain rule is always at play.\[\frac d{dx}(ln(x^2))=\frac{\frac d{dx}(x^2)}{x^2}\to\ \frac{2x}{x^2}\]

Answer 23

what would you say is the derivative of x^3 by chance?

Answer 24

but you said
if we let: a = x^2
ln(a)' = 1/a

Answer 25

if we are deriving "with respect to" x^2, then yes ....

Answer 26

but I guess the how "with respect to" is meant by khan academy isn't formally the same thing as how you defined it?

Answer 27

Is it wrong to think of whatever you are taking respect to as one variable. So in your example if you are taking with respect to x^2 you need to think of it as "a", only one variable, instead of x^2....?

Answer 28

my definitions tend to be based on my own comprehension

Answer 29

Is there no formal definition of this?

Answer 30

im sure there is, its prolly located in a dusty textbook on a shelf someplace :)

Answer 31

x)

Answer 32

I mean I kind of understand it both ways, both in your definition and how khan academy uses it, but that's not getting me anywhere, I need some definition that's always true

Answer 33

the only formal definition of a derivative that I can recall is

\[\lim_{\Delta x\to\ 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\]

Answer 34

yeah, I know that one. But when people say we take derivative of something with respect to something they're just getting me confused because I don't know what they mean, they could mean so many different things...

Answer 35

then lets try this:

what would you say is the derivative of x^3 by chance?

Answer 36

2x^3

Answer 37

uhh
sorry!

Answer 38

ok, now why did you assume it was that?

Answer 39

3x^2
Well, some rule.

Answer 40

3x^2 ... of course :) but your missing something from it

your missing the chain rule part

Answer 41

do we know what this is "with respect to" ?

Answer 42

well not really

Answer 43

then we allow the chain rule to define the unknown

the derivative of: x^3 is, 3x^2 * x'

Answer 44

now, when i say that this is wrt x; what does x' equate to?

Answer 45

wrt?

Answer 46

w.ith
r.espect
t.o

wrt

Answer 47

3x^2 * 1x^(1-1) = 3x^2

Answer 48

if i were to say, what is the derivative of x^3, wrt n; what would we get?

Answer 49

3x^2 * n' ?

Answer 50

not n': x'

we assume by the construct of the question that x is a function of n; or rather x(n)

Dn [x(n)^3] = 3(x(n)^2)*x'(n)

which is usually cleaned up as

Dn [x^3] = 3x^2 * x'

Answer 51

when y is a function of x: y(x) and you take the derivative wrt x;
do you end up with x'? No.

the derivative of y wrt x is y'
the derivative of x wrt n is x'
the derivative of r wrt t is r'

the "wrt" defines the independant variable that we are changing to see how it affect the dependant variable

Answer 52

until we know the "wrt"; the chain rule just pops out a ' version :)

Answer 53

if you have an example we can run thru, something that your having issues on; that might be more productive

Answer 54

I'm not sure if I'm getting it but I'll reread it in a while, though you'll have to excuse me because I have to go afk for a while

Answer 55

thanks for the help so far though

Answer 56

ill be in class from 130 to 5 or so, so i might not get back to this till 2mrw; but im sure there are others capable of help you as well

Answer 57

ok :)

Answer 58

I'm still not getting this wrt... Yes, I reread it

Answer 59

Dn [x^3] = 3x^2 * x'

What is x' here?

Answer 60

And Dn?

Answer 61

derivative with respect to n?

Answer 62

that n makes no sense to me either

Answer 63

oh glad I'm not alone

Answer 64

usually you may see\[D_x[f(x)]=f'(x)\]which is the derivative with respect to x

Answer 65

\[= \frac{dy}{dx}f(x) ?\]

Answer 66

you could also see\[\frac{d^n}{dx^n}f(x)=f^{(n)}(x)\]which would mean the \(n^{th}\) derivative with respect to x

Answer 67

what you wrote right now makes little sense

Answer 68

usually, in single-variable calculus, y is just a variable used to represent a function of x f(x)

Answer 69

so\[\frac{dy}{dx}=\frac d{dx}f(x)=f'(x)\]

Answer 70

how do you read the second one?

Answer 71

difference in ? over difference in x

Answer 72

so what the heck is\[\frac{dy}{dx}f(x)=\frac d{dx}[f(x)]f(x)=f'(x)f(x)\]I guess... this is an ill- formed expression
how to read it? all these things are read the derivative with respect to (whatever is bottom,
d/dx is wrt x, d/dy is wrt y, d/d(purple) is the derivative with respect to purple, etc.)

Answer 73

\[\frac {d^n}{dx^n}f(x)=f^{(n)}(x) \]is read as "the \(n^{th}\) derivative with respect to x of f of x"

Answer 74

all right. But I still don't get this with respect to. Can you introduce me to a formal definition?

Answer 75

formal... hm...

Answer 76

But wait, shouldn't that last latex expression be read as "the nth derivative with respect to x^n of f(x)" ?

Answer 77

no it is not, don't let old algebra notation confuse you
d/dx is not a fraction in calculus it is an operator, and the x^n has nothing to do with x raised to the n power

how about if I say this
let's say y is a function of x