Question

Find the equation of the tangent to the curve `x + xy + y = 5` at `x = 5` help?

Answer 1

`y = 5/6 - 1/6x`

Explanation:

Start by finding the y-value:

`5 + 5y + y = 5`

`6y = 0`

`y = 0`

Now we find the derivative using implicit differentiation.

`1 + y + x(dy/dx) + dy/dx = 0`

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

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

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

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

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

At `(5, 0)`, the derivative will have value

`dy/dx= -1/6`

Now use point-slope form to find the equation:

`y - y_1 = m(x - x_1)`

`y - 0 = -1/6(x - 5)`

`y = -1/6x +5/6`

Hopefully this helps!

Answer 2

`y = -1/6x+5/6`

Explanation:

Begin by finding the y coordinate at `x = 5`:

`5+5y+y=5`

`5y+y = 0`

`6y = 0`

`y = 0`

The point of tangency is `(5,0)`

Compute the first derivative of the curve:

`1+y+xdy/dx+dy/dx =0`

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

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

The slope, m, of the tangent line is the first derivative evaluated at the point `(5,0)`:

`m = -(0+1)/(5+1)`

`m = -1/6`

Use the point-slope form for the equation of a line:

`y = m(x-x_0)+y_0`

`y = -1/6(x-5)+0`

`y = -1/6x+5/6`

Here is the a graph of the curve and the tangent line: