Using the first principle, find partial f/ partial x and partial f / partial y (i)f(x,y)=x^3y-xy^3. (ii)f(x,y)=[(x^2y+2xy^2)/(x^2+y)].Thanks?

f(x,y) = x³y - xy³

∂f(x,y)/∂x = fx = lim h-->0 { [ f(x+h),y - f(x,y) ]/h } , or if you like h = Δx

f((x+h),y) - f(x,y) =
(x+h)³y - (x+h)y³ - (x³y - xy³)

(x³+3x²h+3xh²+h³)y - (x+h)y³ - (x³y - xy³)
x³y+3x²yh+3xyh²+yh³-xy³-y³h-x³y+xy³ , simplify
(3x²y - y³)h + 3xyh² + yh³ , div by h =>
(3x²y - y³) + 3xyh + yh² , let h-->0

fx = 3x²y - y³
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same procedure for fy , which I leave to you!

i)
f(x,y) = x^3y - xy^3
when we partially differentiate with respect to x, y is a constant
f( x+h, y) - f(x, y) = ( (x+h)^3 y - (x+h) y^3 ) - (x^3y - xy^3)
f( x+h, y) - f(x, y) = ( x^3y+3x^2hy +3xh^2y+h^3y - xy^3-hy^3 -x^3y+xy^3)
f( x+h, y) - f(x, y) = ( 3x^2hy +3xh^2y+h^3y -hy^3 )
[f( x+h, y) - f(x, y)] / h = ( 3x^2y +3xhy+h^3y - y^3 )

lim h-->0 [f( x+h, y) - f(x, y)] / h = lim h-->0 ( 3x^2y +3xhy+h^3y - y^3 )
= 3x^2y -y^3

∂f/∂x = 3x^2y - y^3


f(x,y) = x^3y - xy^3
when we partially differentiate with respect to y, x is a constant
f(x, y+h) - f(x,y) = ( x^3(y+h) - x(y+h)^3 ) - (x^3y -xy^3)
f(x, y+h) - f(x,y) = ( x^3y+x^3h - (y^3x+3y^2hx+3yh^2x+h^3x) ) - (x^3y -xy^3)
f(x, y+h) - f(x,y) = ( x^3y+x^3h - y^3x-3y^2hx-3yh^2x-h^3x - x^3y +xy^3)
f(x, y+h) - f(x,y) = (x^3h -3y^2hx-3yh^2x-h^3x )
[(f(x,y+h) -f(x,y) ] / h = (x^3 -3y^2x-3yhx+h^2 x)

lim h-->0 [(f(x,y+h) -f(x,y) ] / h = lim h-->0 (x^3 -3y^2x-3yhx+h^2 x)
= x^3 - 3y^2 x
∂f/∂y = x^3 - 3x y^2