Question

How do you use implicit differentiation to find dy/dx given `y=(x+y)^2`?

Answer 1

I got

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

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Since `y` is a function of `x` (either `y` in the equation), the derivative with respect to `x` must include `(dy)/(dx)` whenever the derivative of `y` with respect to `x` is taken.

`d/(dx)[y(x)] = d/(dx)[(x+y)^2]`

`(dy)/(dx) = 2(x+y)^(1) * stackrel("Chain Rule")overbrace(d/(dx)[x+y])`

`= 2(x+y)(1 + (dy)/(dx))`

`= (2x + 2y)(1 + (dy)/(dx))`

`= 2x + 2x (dy)/(dx) + 2y + 2y(dy)/(dx)`

Now isolate the similar terms.

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

`(dy)/(dx)[1 - 2x - 2y] = 2x + 2y`

`=> color(blue)((dy)/(dx) = (2x + 2y)/(1 - 2x - 2y))`

Answer 2

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

Explanation:

Here is an alternative answer.

Expand inside the parentheses:

`y = (x + y)(x + y)`

`y = x^2 + xy + xy + y^2`

`y = x^2 + 2xy + y^2`

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

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

`1(dy/dx) = 2x + 2y + 2x(dy/dx) + 2y(dy/dx)`

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

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

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

Hopefully this helps!