Find partial derivatives ∂f/∂x and ∂f/∂y of the following function:?

Question

f(x,y) = (2x - y) / (x + 3y)

Answer 1

∂f/∂x = [(x + 3y)*2 - (2x-y)*1]/(x+3y)^2
=7y/(x+3y)^2

∂f/∂y = [(x+3y)*-1 - (2x-y)*3]/(x+3y)^2
=-7x/(x+3y)^2

Answer 2

since f(x, y) = (2x - y) / (x + 3y)

letting y in f(x, y) a constant...

f(x, y) can be re-written as...

f(x, y) = 2 - 7y/(x + 3y)

so...

∂f/∂x = 7y / ((x + 3y)^2)

to get ∂f/∂y, easier... do the same process, remembering to set x as a constant... so...

f(x,y) = (2x - y) / (x + 3y) can be written as...

f(x, y) = -1/3 + 7/3(x/(x+3y))

∂f/∂y =-3*(7/3(x/(x+3y)^2))

∂f/∂y = -7x / (x+3y)^2

Answer 3

1) Generally, ∂f/∂x is the same as df/dx where y is treated as a constant. So, you do both derivatives the same way

2) Let f(x,y) = N(x,y) / D(xy)

3) ∂f/∂x = ( D * ∂N/∂x - N * ∂D/∂x ) / D^2 = 7y / ( ( x + 3y )^2 )

4) ∂f/∂y = ( D * ∂N/∂y - N * ∂D/∂y ) / D^2 = -7x / ( ( x + 3y )^2 )

Answer 4

∂f / ∂x = [ (x + 3y).(2) - (2x - y).1 ] / (x + 3y)²
∂f /∂x = (2x + 6y - 2x + y) / (x + 3y)²
∂f / ∂x = 7y / (x + 3y)²

∂f / ∂y = [- (x + 3y) - 3(2x - y) ] / (x + 3y)²
∂f / ∂y = (- 7x) / (x + 3y)²

Answer 5

∂f/∂x = [2(x + 3y) - (2x -y)]/(x +3y)^2 = 5y/(x +3y)^2

∂f/∂y = [(x -3y)(-1) - (2x - y)(-3)]/(x +3y)^2 = 5x/(x +3y)^2

Answer 6

fx= 2xy/yx^2+yexp(xy) = 2/x + yexp(xy) fy= x^2/yx^2 +xexp(xy) = a million/y + xexp(xy) fxx= -a million/x^2 + y^2exp(xy) fxy= exp(xy)+xyexp(xy) fyy= -2/y^2 + x^2exp(xy) fyx= exp(xy)+xyexp(xy)