If dy/dx=x^2y, how do i get the 2nd derivative? (1 Viewer)

[vanbasten]

dy/dx=x^2y

d^2y/dx^2=?

 

[withoutaface]

d^2y/dx^2=2xy+x^2dy/dx
=2xy+x^4y

But this is 4u, not 3u.

 

[utopia_parkway]

could you explain to me how you got that answer, its just not working for me. did you use implicid diffeneration and the chain rule?

 

[Supra]

i think hte question is dy/dx = x²y.. find the second derivative...coz intitially i thought it was x^(2y) but no thats jus fukd...so neways yes its implicit and product rule

u=x²
u'=2x
v=y
v'=dy/dx
chain rule says vu'+uv'
so then u get:
d²y/dx²= 2xy + x²dy/dx and so on

btw this is most probably 4u work i dont know if implicit differentiation is in the 3u course

EDIT: whoops fixed

 

[Slidey]

coz intitially i thought it was x^(2y) but no thats jus fukd

let z=x^(2y)
lnz=2ylnx
(lnz)'=(2ylnx)'
z'/z=2y'/x
z'=2zy'/x
(x^(2y))'=2.(y'/x).x^(2y)

d/dx[x^(2y)]=2y'.x^(2y-1)

Oh, and vu'+uv' is product rule, not chain rule.

 

[utopia_parkway]

cooiles, i thought it was x^(2y) as well. x^2.y is much easier to do

 

[Rorix]

y' = x^(2y)
lny' = 2y ln x
1/y' * y'' = 2y/x + 2lnx
y'' = 2yy'/x + 2y'lnx
y'' = 2yx^(2y-1) + 2x^(2y)lnx

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