what is d^2y/dx^2 of (x^2)(y^2)=9?

Answer 1

are you asking for the derivative of dy squared with respect to x squared when x squared times y squared equals 9?

Because your notation is confusing

Answer 2

(x^2)(y^2)=9

the first deriv is
2x y^2 +x^2 2y dy/dx=0
arrange it dy/dx =-2yxy^2/(2yx^2) cancle out
u get dy/dx = -y/x
now u need the secod deriv

d^2y/dx^2 = ((-dy/dx x ) - (-y))/x^2

as we know the value of dy/dx =-y/x so subtitute in the second deriva

u get
d^2y/dx^2 = ((y/x)x+ y)/x^2 ==>
=> d^2y/dx^2 =2y/x^2

Answer 3

y^2=9/x^2
on differentiating with respect to x
2y(dy/dx)=9*(-2)(1/x^3)=-18/x^3
dy/dx=18/(2x^3y)=9/(x^3 y)
d^2y/dx^2=(3x^2+x^3 dy/dx) / (x^6 y^2)........2
substituting dy/dx in 2
d^2y/dx^2=(3x^2 y+9) / (x^6 y ^2)
tats it !!!!!!!!!!!

Answer 4

i think your notation is fine. easiest to do it this way:

rewrite as:

y^2=9x^(-2)
2y*dy/dx=-18x^(-3) - differentiating once
2y*(d^2y/dx^2)+ 2(dy/dx) * (dy/dx) = 54x^(-4)
so, grouping terms,

2y(d^2y/dx^2) + 2[dy/dx]^2 - 54x^(-4) = 0
i.e. (d^2y/dx^2) + [dy/dx]^2 - 27x^(-4) = 0
you can rearrange to put d^2y/dx^2 = something if u like

Answer 5

y^2=9x^-2
so y= 3x^-1
differentiating
dy/dx = -3x^-2
once more differntiating
d^2y/dx^2 = 6x^-3

Answer 6

(2*y^2) + (2*x^2) + 8*x*y

Answer 7

characters in a cloud for swearing, in a comic strip.

Answer 8

its bollocks