# Use implicit differentiation to find an eqution of the tangent line of to the graph at the given point. `x^(2)+xy+y^(2) = 4, (4,0)` You need to use implicit differentiation, such that:

`2x + y + x*(dy)/(dx) + 2y*(dy)/(dx) = 0`

You need to isolate the terms that contain `(dy)/(dx)` to the left side, such that:

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

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

You need to...

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You need to use implicit differentiation, such that:

`2x + y + x*(dy)/(dx) + 2y*(dy)/(dx) = 0`

You need to isolate the terms that contain `(dy)/(dx)` to the left side, such that:

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

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

You need to evaluate the equation of the tangent line to the graph of the function, at (4,0), such that:

`y - 0 = (dy)/(dx)|_(4,0)(x - 4)`

`y = -(2x + y)/(x + 2y)|_(4,0)(x - 4)`

`y = -(2*4 + 0)/(4 + 2*0)(x - 4)`

`y = -2(x - 4) => y = -2x + 8`

Hence, evaluating the equation of the tangent line to the graph, at the point `(4,0)` , using implicit differentiation, yields ` y = -2x + 8.`

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