Question

How do you find `(df)/dy` and `(df)/dx` of `f(x,y)=(3x^2-2e^-y)/(2e^x+y^-3)`, using the quotient rule?

Answer 1

`(partial f)/(partial x) = ( 6x(2e^x+y^-3) - 2e^x(3x^2-2e^-y) ) / (2e^x+y^-3)^2`

`(partial f)/(partial y) = ( 2e^-y(2e^x+y^-3) +3y^-4(3x^2-2e^-y) )/ (2e^x+y^-3)^2`

Explanation:

The quotient rule for multi variable calculus is identical to single variable calculus except that we perform the differentiation wrt to a specific variable whilst holding other variable constant. so:

`(partial)/(partial x) (u(x,y))/(v(x,y))= (v(partial u)/(partial x) - u(partial u)/(partial x)) / v^2`

`(partial)/(partial y) (u(x,y))/(v(x,y))= (v(partial u)/(partial y) - u(partial u)/(partial y)) / v^2`

We have:

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

If we differentiate wrt `x` using the quotient rule (whilst holding `y` constant) we get:

`(partial f)/(partial x) = ( (2e^x+y^-3)(6x) -(3x^2-2e^-y) (2e^x) ) / (2e^x+y^-3)^2`
`\ \ \ \ \ \ \ \= ( 6x(2e^x+y^-3) - 2e^x(3x^2-2e^-y) ) / (2e^x+y^-3)^2`

If we differentiate wrt `y` using the quotient rule (whilst holding `x` constant) we get:

`(partial f)/(partial y) = ( (2e^x+y^-3)(2e^-y) -(3x^2-2e^-y) (-3y^-4) ) / (2e^x+y^-3)^2`
`\ \ \ \ \ \ \= ( 2e^-y(2e^x+y^-3) +3y^-4(3x^2-2e^-y) )/ (2e^x+y^-3)^2`