Question

How do you implicitly differentiate `y= xy^2 + x ^2 e^(x -2y)`?

Answer 1

`dy/dx=(y^2+x^2(e^(x-2y))+2xe^(x-2y))/(1-2xy+2x^2e^(x-2y))`

Explanation:

Take the derivative of the equation except for every time you differentiate `y`, it becomes `dy/dx`. You will also have to use the product rule.

Note: `e^(x-2y)=(e^x)(e^-(2y))`. So the equation is actually:

`y=xy^2+x^2(e^x)(e^-(2y))` so you will have to use the product rule twice for the second term:

`dy/dx=x(2ydy/dx)+y^2(1)+(x^2(e^xe^-(2y))' + (e^x)(e^-(2y))(2x)`

Use the product rule again:

`dy/dx=2xy(dy/dx)+y^2+(x^2)((e^x)(-2e^(-2y)dy/dx)+(e^-2y)(e^x))+(e^x)(e^-(2y))(2x)`

Distribute the x^2:

`dy/dx=2xy(dy/dx)+y^2+x^2(e^x)(-2e^(-2y)dy/dx)+(x^2)(e^(-2y))(e^x)+(e^x)(e^-(2y))(2x)`

Separate the terms with `dy/dx`

`dy/dx-2xydy/dx+2x^2e^xe^-(2y)(dy/dx)=y^2+x^2(e^(-2y))(e^x)+2x(e^x)(e^(-2y))`

Factor out the `dy/dx`:

`(dy/dx)(1-2xy+2x^2e^(x-2y))=y^2+x^2(e^(x-2y))+2xe^(x-2y)`

Divide both sides by `(1-2xy+2x^2e^(x-2y))`:

`dy/dx=(y^2+x^2(e^(x-2y))+2xe^(x-2y))/(1-2xy+2x^2e^(x-2y))`

Answer 2

`d/dx( y= xy^2 + x ^2 e^(x -2y))`

`implies y' = y^2 + 2 x y y' + 2 x e^(x -2y) + x^2 (color(red)(1)- 2y') e^(x -2y)`

`implies y'(1- 2 x y + 2x^2 e^(x -2y) ) = y^2 + (color(red)(x^2 +) 2 x) e^(x -2y)`

`implies y'= ( y^2 + (color(red)(x^2 +) 2 x) e^(x -2y) )/(1- 2 x y + 2x^2 e^(x -2y) )`

Answer 3

`dy/dx = (y^2+(x^2+2x)e^(x-2y))/(1+2x^2e^(x-2y)-2xy)`

Explanation:

Using the Implicit Function Theorem we have:

`dy/dx = -((partial f)/dx)/((partial f)/dy)`

Where:

`f(x,y) = 0`

We have:

`\ \ \ \ \ y=xy^2 + x^2e^(x-2y)`
`:. y=xy^2 + x^2e^x e^(-2y)`

Let:

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

Then the partial derivatives are:

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

And:

`(partial f)/(partial y) = 2xy-2x^2e^x e^(-2y)-1`
`\ \ \ \ \ \ \ = 2xy-2x^2e^(x-2y)-1`

And so:

`dy/dx = -(y^2+(x^2+2x)e^(x-2y))/(2xy-2x^2e^(x-2y)-1)`

`\ \ \ \ \ \ = (y^2+(x^2+2x)e^(x-2y))/(1+2x^2e^(x-2y)-2xy)`