Can someone help me find the second derivative of this function?

Question

x squared y squared - 2x=3

x^2 y^2 - 2x = 3

can u plz give me the simplified answer

Answer 1

x^2 y^2 - 2x = 3

-First derivative ( dy/dx )
by using the Product Rule

[(x^2)(2 dy/dx) + (y^2)(2x)] - 2 = 0
[(x^2)(2 dy/dx) + (y^2)(2x)] = 2
(x^2)(2 dy/dx) = 2 - (y^2)(2x)
2 dy/dx = 2 - (y^2)(2x) / x^2
dy/dx =[ 2 - (y^2)(2x) / x^2 ] / 2
dy/dx = 2 - y^2 (2x) / 2x^2
dy/dx = 2 - y^2 / x

-Then, the second derivate ( d^2y/dx^2 )
by using the Quotient Rule

dy/dx = (x - 2dy^2/dx^2 - 1 +y^2) / (x^2)
2 dy^2/dx^2 = x - 1 + y^2 - (dy/dx)
dy^2/dx^2 = (x - 1 + y^2 - dx/dx) / 2

Answer 2

We need a bit of implicit differentiation here, and the use of the product rule.
For a function f(y)
d[f(y)]/dx = d[f(y)]/dy . (dy/dx)
Now for (xy)^2 - 2x = 3
differentiating,
2(xy).{x(dy/dx) + y} - 2 = 0
then, dy/dx = (1/x^2y) - (y/x)
now again repeat the process and substitute for dy/dx where necessary. You wont get a beautiful answer, but simplify it down.
Else this can be approached without implicity as,
y = {(3+2x)/x^2}^(1/2) = V(3+2x) / x
use the quotient rule twice. V stands for square root.

Answer 3

I assume you're looking for d^2 y/dx^2.

Differentiate x^2 y^2 - 2x = 3 implicitly with respect to x:

x^2 2y dy/dx + 2x y^2 - 2 = 0

Solve for dy/dx:

x^2 2y dy/dx = 2 - 2x y^2, so
dy/dx = (2-2xy^2) / (2x^2 y)
= (1-xy^2) / (x^2 y).

Now differentiate both sides with respect to x again:

d^2 y/dx^2 = [(x^2 y)(-y^2-2xy dy/dx) - (1-xy^2)(2xy +x^2 dy/dx)] / (x^2 y)^2.

To complete this, substitute the above expression for dy/dx into this second derivative, and simplify.

Answer 4

Hey hey! I hope this helps.. I tried typing it out with the right symbols in word but it didnt come out properly here.

x^2*y^2 – 2x = 3
y2 = (3 + 2x)x^-2
y = (3x^-2 + 2x^-1)^(1/2)

dy/du = ½(u)
u = 3x^-2 + 2x^-1
du/dx = -6x^-3 – 2x^-2

dy/dx = dy/du*du/dx
dy/dx = ½(3x^-2 + 2x^-1)( -6x^-3 – 2x^-2)

thats the first derivative. I'm sorry i give up.. im on holiday i cant think.

Answer 5

first find the first derivative..then do it again! and tah-dah! you have the second derivative. you have to use the multiplication rule for the x^2y^2 one, but other than that its really easy. maybe you should check your book for an example.

Answer 6

i suppose ur equation is x^2(y^2)-2x=3

first, finding dy/dx:

dy/dx: 2x(y^2)+2y(dy/dx)(x^2)-2=0 (by product rule and differentiating y implictly w.r.t. x)

then d^2y/dx^2:

d^2y/dx^2: 2y^2+2(d^2y/dx^2)(x^2)=0 (differentiating again)

y^2+(d^2y/dx^2)(x^2)=0 (dividing both sides by 2 to remove unnecessary coefficient)

(d^2y/dx^2)(x^2)= -y^2 (moving y^2 to the other side of equation)

Hence: d^2y/dx^2 = (-y^2)/(x^2) (dividing by x^2)

Hope its right ~.~

Answer 7

x^2y^2 - 2x = 3
or x(xy^2 - 2)=3