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second derivative help needed

how would i find the second derivative of y=vx (differentiatate with respect to x)
please show full steps.
i found the first derivative using implicit diff
y=vx
y'=x(dv/dx)+v
y''=????
i think what confuses me is x(dv/dx) how would i diff that????
@RDKgames
Original post by assassinbunny123
how would i find the second derivative of y=vx (differentiatate with respect to x)
please show full steps.
i found the first derivative using implicit diff
y=vx
y'=x(dv/dx)+v
y''=????
i think what confuses me is x(dv/dx) how would i diff that????
@RDKgames



@ghostwalker
Original post by assassinbunny123

@RDKGames
Use product rule again, and if you differentiate dv/dx (i.e. d/dx of dv/dx) you get d2vdx2\frac{d^2 v}{dx^2} (the second derivative).
Original post by assassinbunny123

i think what confuses me is x(dv/dx) how would i diff that????
@RDKgames


Product rule again.
Reply 5
Avatar for LeonDH
LeonDH
9
Original post by assassinbunny123
how would i find the second derivative of y=vx (differentiatate with respect to x)
please show full steps.
i found the first derivative using implicit diff
y=vx
y'=x(dv/dx)+v
y''=????
i think what confuses me is x(dv/dx) how would i diff that????
@RDKgames


After you differentiate with respect to x first time, you are left with v, which is a constant, so you can not differentiate again. It's like trying to differentiate a number like 4.
Original post by assassinbunny123
how would i find the second derivative of y=vx (differentiatate with respect to x)
please show full steps.
i found the first derivative using implicit diff
y=vx
y'=x(dv/dx)+v
y''=????
i think what confuses me is x(dv/dx) how would i diff that????
@RDKgames


https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-advanced/ab-second-derivatives/v/second-derivatives
This might help...
Just do it in the normal way

y' = xv' +.v
(xv')' = (x)'(v') + (x)(v')'
= (dv/dx) + x(d²v/dx²)
(edited 5 years ago)
Original post by LeonDH
After you differentiate with respect to x first time, you are left with v, which is a constant, so you can not differentiate again. It's like trying to differentiate a number like 4.


vv is not necessarily a constant.

Also, if it was, what's the issue with differentiating it??? Second derivative would just be zero.
(edited 5 years ago)
Original post by assassinbunny123
not quite
y'=x(dv/dx)+v
y''=x(d2v/dx2) + dv/dx+dv/dx
=x(d2v/dx2)+2(dv/dx)


Correct, though they said that (xv′)′=dvdx+xd2vdx2(xv' ) ' = \dfrac{dv}{dx} + x \dfrac{d^2 v}{dx^2} and not that (y′)′=dvdx+xd2vdx2(y' ) ' = \dfrac{dv}{dx} + x \dfrac{d^2 v}{dx^2} which is what it seems you incorrectly saw.
Their post is fine!
(edited 5 years ago)
Original post by 3pointonefour
Just do it in the normal way

y' = xv' +.v
(xv':wink:' = (x)'(v':wink: + (x)(v':wink:'
= (dv/dx) + x(d²v/dx²)

misread your solutions i think the winking faces put me off lol.
Original post by assassinbunny123
misread your solutions i think the winking faces put me off lol.


Yeah sorry that was TSR's fault. Fixed now.

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