find first and second partial derivative of f(x,y) = x^2-3x/1+x^2y^2

Question

find the first and second partial derivative of the question

Answer 1

Let u=x²-3x, then ∂u/∂x=2x-3, ∂u/∂y=∂²u/∂y²=0;

Let v=(1+x²y²)^-n, then ∂v/∂x=-n(2xy²)(1+x²y²)^-n-1, ∂v/∂y=-n(2x²y)(1+x²y²)^-n-1.

These derivatives may be useful as we progress through the 1st and 2nd derivatives:

 ∂f/∂x, ∂f/∂y, (∂/∂x)(∂f/∂y), (∂/∂y)(∂f/∂x), ∂²f/∂x², ∂²f/∂y².

We would expect the 3rd and 4th of these listed derivatives to be equal.

∂f/∂x=∂(uv)/∂x=v∂u/∂x+u∂v/∂x=(1+x²y²)^-1(2x-3)-(x²-3x)(2xy²)(1+x²y²)^-2,

∂f/∂x=(1+x²y²)^-1(2x-3)-(2x³y²-6x²y²)(1+x²y²)^-2,

∂f/∂x=[(2x-3)(1+x²y²)-(2x³y²-6x²y²)](1+x²y²)^-2,

∂f/∂x=(2x+2x³y²-3-3x²y²-2x³y²+6x²y²)(1+x²y²)^-2,

∂f/∂x=(3x²y²+2x-3)(1+x²y²)^-2.

Let u=Let u=x²-3x, then ∂u/∂x=2x-3, ∂u/∂y=∂²u/∂y²=0; ∂²u/∂x²=2.

Let v=(1+x²y²)^-n, then ∂v/∂x=-n(2xy²)(1+x²y²)^-n-1, ∂v/∂y=-n(2x²y)(1+x²y²)^-n-1.

These derivatives may be useful as we progress through the 1st and 2nd derivatives:

∂f/∂x, ∂f/∂y, (∂/∂x)(∂f/∂y), (∂/∂y)(∂f/∂x), ∂²f/∂x², ∂²f/∂y².

We would expect the 3rd and 4th of these listed derivatives to be equal.

∂f/∂x=∂(uv)/∂x=v∂u/∂x+u∂v/∂x=(1+x²y²)^-1(2x-3)-(x²-3x)(2xy²)(1+x²y²)^-2,

∂f/∂x=(1+x²y²)^-1(2x-3)-(2x³y²-6x²y²)(1+x²y²)^-2,

∂f/∂x=[(2x-3)(1+x²y²)-(2x³y²-6x²y²)](1+x²y²)^-2,

∂f/∂x=(2x+2x³y²-3-3x²y²-2x³y²+6x²y²)(1+x²y²)^-2,

∂f/∂x=(3x²y²+2x-3)(1+x²y²)^-2.

∂f/∂y=∂(uv)/∂y=v∂u/∂y+u∂v/∂y=0-(2x²y)(x²-3x)(1+x²y²)^-2=(-2x⁴y+6x³y)(1+x²y²)^-2.

Let u=3x²y²+2x-3, then ∂u/∂y=6x²y; v=(1+x²y²)^-2.

(∂/∂y)(∂f/∂x)=∂(uv)/∂y=v∂u/∂y+u∂v/∂y,

(∂/∂y)(∂f/∂x)=6x²y(1+x²y²)^-2+(3x²y²+2x-3)(-2(2x²y)(1+x²y²)^-3,

(∂/∂y)(∂f/∂x)=[(6x²y(1+x²y²)-4x²y(3x²y²+2x-3)](1+x²y²)^-3,

(∂/∂y)(∂f/∂x)=(6x²y+6x⁴y³-12x⁴y³-8x³y+12x²y)(1+x²y²)^-3,

(∂/∂y)(∂f/∂x)=(-6x⁴y³-8x³y+18x²y)(1+x²y²)^-3.

∂f/∂y=(-2x⁴y+6x³y)(1+x²y²)^-2; let u=-2x⁴y+6x³y, then ∂u/∂x=-8x³y+18x²y; v=(1+x²y²)^-2;

(∂/∂x)(∂f/∂y)=∂(uv)/∂x=v∂u/∂x+u∂v/∂x,

(∂/∂x)(∂f/∂y)=(1+x²y²)^-2(-8x³y+18x²y)+(-2x⁴y+6x³y)(-2(2xy²)(1+x²y²)^-3,

(∂/∂x)(∂f/∂y)=[(-8x³y+18x²y)(1+x²y²)-4xy²(-2x⁴y+6x³y)](1+x²y²)^-3,

(∂/∂x)(∂f/∂y)=(-8x³y+18x²y-8x⁵y³+18x⁴y³+8x⁵y³-24x⁴y³)(1+x²y²)^-3,

(∂/∂x)(∂f/∂y)=(-6x⁴y³-8x³y+18x²y)(1+x²y²)^-3=(∂/∂y)(∂f/∂x), confirming expectation.

Let u=3x²y²+2x-3, then ∂u/∂x=6xy²+2; v=(1+x²y²)^-2.

∂²f/∂x²=v∂u/∂x+u∂v/∂x=(6xy²+2)(1+x²y²)^-2-4xy²(3x²y²+2x-3)(1+x²y²)^-3,

∂²f/∂x²=[(6xy²+2)(1+x²y²)-4xy²(3x²y²+2x-3)](1+x²y²)^-3,

∂²f/∂x²=(6xy²+2+6x³y⁴+2x²y²-12x³y⁴-8x²y²+12xy²)(1+x²y²)^-3,

∂²f/∂x²=(-6x³y⁴-6x²y²+18xy²+2)(1+x²y²)^-3.

Let u=-2x⁴y+6x³y, then ∂u/∂y=-2x⁴+6x³; v=(1+x²y²)^-2.

∂²f/∂y²=v∂u/∂y+u∂v/∂y=(1+x²y²)^-2(-2x⁴+6x³)-4x²y(-2x⁴y+6x³y)(1+x²y²)^-3,

∂²f/∂y²=[(1+x²y²)(-2x⁴+6x³)-4x²y(-2x⁴y+6x³y)](1+x²y²)^-3,

∂²f/∂y²=(-2x⁴+6x³-2x⁶y²+6x⁵y²+8x⁶y²-24x⁵y²)(1+x²y²)^-3,

∂²f/∂y²=(6x⁶y²-18x⁵y²-2x⁴+6x³)(1+x²y²)^-3.