Question

How to determine the equation of the tangent line to `y^3+xy=22` at `(7,2)`?

Answer 1

Implicitly differentiate the given equation.
Obtain the slope by evaluating the first derivative at the given point.
Use the point-slope form of the equation of a line to write the equation.

Explanation:

Verify that the point lies `(7,2)` on the curve `y^3+xy=22`:

`2^3+(7)(2)=22`

`8 + 14 = 22`

`22 = 22 larr` verified.

Implicitly differentiate `y^3+xy=22`:

`(d(y^3))/dx+ (d(xy))/dx = (d(22))/dx`

The first term requires the use of the chain rule:

`(d(y^3))/dx = (d(y^3))/dy dy/dx`

`(d(y^3))/dx = 3y^2 dy/dx`

Substitute the above into the equation:

`3y^2 dy/dx+ (d(xy))/dx = (d(22))/dx`

The next term requires the use of the product rule:

`(d(x y))/dx = dx/dxy+xdy/dx`

`(d(x y))/dx = y+xdy/dx`

Substitute the above into the equation:

`3y^2 dy/dx+ y+xdy/dx = (d(22))/dx`

The last term is the derivative of a constant:

`3y^2 dy/dx+ y+xdy/dx = 0`

Move the term that does not contain `dy/dx` to the right:

`3y^2 dy/dx+xdy/dx = -y`

Factor out `dy/dx`:

`(3y^2+x)dy/dx = -y`

Divide both sides by the leading coefficient:

`dy/dx = -y/(3y^2+x)`

The slope, `m`, of the tangent line, is the first derivative evaluated that the point of tangency, `(7,2)`:

`m = -(2)/(3(2)^2+7)`

`m = -2/19`

Begin the final step with the point-slope form of the equation of a line:

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

Substitute `m = -2/19, x_0 = 7, and y_0 = 2`

`y = -2/19(x-7)+2`