Question

If I am finding dy/dx it is the derivative of y/derivative of x, d^2y/dx^2 it is derivative of (dy/dx)/derivative of x, and d^3y/dx^3 would be derivative of (d^2y/dx^2)/derivative of x and so on and so on? keeping the denominator the derivative of x each time?

Answer 1

dy/dx means the derivative of y with respect to x.

Answer 2

And the correct Liebnitz notation is:

First derivative: \[\frac{ dy }{ dx }\]
Second derivative: \[\frac{ d^2y }{ dx^2 }\]
Third derivative: \[\frac{ d^3y }{ dx^3 }\]

Answer 3

but when working out the third derivative there on top would be the derifvative of the second derivative / by the derivative of the original x, yes?

Answer 4

Third derivative of y is:
\[\frac{ d }{ dx }(\frac{ d }{ dx }(\frac{ dy }{ dx }))\]

Is this what you're asking?

Answer 5

it just looks like each time you take the derivative of the one before and divide it by the derivative of the orginal x, like the d^3y = (d^2y/dx^2)/(dx/dt)

Answer 6

Don't think of /dx as division! It's a notation, not a ratio.

Answer 7

Want to try an easy example?

Answer 8

yeah bud im down

Answer 9

\[y = x^6\]
\[\frac{ dy }{ dx } = 6x^5\]
\[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }(\frac{ dy }{ dx }) = 30x^4\]

Answer 10

Are you following so far?

Answer 11

yes sir

Answer 12

Cool. Try finding the third derivative.

Answer 13

120x^3 right?

Answer 14

yeap!

Answer 15

but what if it is y= something and x=something else

Answer 16

I'll make some more examples and you can try them :)

Answer 17

like y=t^2) and x= t^4

Answer 18

thanks for all the help my friend, got a cal 2 test in the morning and trying to sure things up

Answer 19

try this:
\[\frac{ d }{ dr }(r^3 + 2r^2)\]

Answer 20

Wait so like:
\[\frac{ d(t^2) }{ d(t^4) } ???\]

Answer 21

3r^2 + 6r^2, but i am trying to find it when a given is y= and a x=

Answer 22

differectial and parametric form it says

Answer 23

I don't understand. You're trying to differentiate with respect to a function?

Answer 24

i'm not sure of the technical way of saying it but asks for the parametric form of a third dericative, it gives a y= and a x= then for the second derivative you take the derivative of y/derivative of x then for the third i think it is the derviative of the second / the derivative of x=

Answer 25

it not just the derivative of a y= but how to find a parametric formula for the second or third dericative when given a y= and a x=

Answer 26

Is it perhaps: x = f(t), y = g(t)?

And they want dy/dx?

Ifso, you use the chain rule: (dy/dt)/(dx/dt)

Answer 27

the questions read similar to this: For the parametric equations x=t +1 and y= t^2 +3t, find (d^2y)/(dx^2)

Answer 28

Note that the higher order derivatives aren't so simple.

\[\frac{ d^2y }{ dx^2 } = \frac{ \frac{ d }{ dt }(\frac{ dy }{ dx }) }{ \frac{ dx }{ dt } }\]

Answer 29

looking at the notation and working backwards it looks like the numerator changes each tome to the next derivative while the denomenator stays the same no matter what degree you derive to?

Answer 30

yeah thats it, to the third is it not the answer to that derived/ derivative of x=?

Answer 31

I might not have to do to the third I have not checked, but I couldnt find anything in the book about it

Answer 32

im on some mirc channels too, I might swing in there and ask too. I appreciate all the help with this and I am very thankfull for your time my friend. Thank you again, and have a good one.

Answer 33

It's a pain to type but this site has a definition of the third derivative: http://17calculus.com/calc06-parametrics-calculus.php

Answer 34

sweet thank you my friend

Answer 35

Glad to help. Cheers!