Find the slope of the tangent line to the graph of the equation xy+12=0 at (-2, 6)

Question

I got m = -3

Answer 1

Instead of solving the equation xy + 12 = 0 for y, let's just do an implicit differentiation.

We'll take the derivative with respect to x (i.e., the d/dx) of both sides of the equation:

(d/dx)(xy + 12) = (d/dx)0

Therefore

(d/dx)(xy) + (d/dx)12 = (d/dx)0

Since the derivative of any constant is zero, this simplifies to

(d/dx)(xy) = 0

Now we use the rule for "derivative of a product". The product in this case is x times y:

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

Since dx/dx =1, this simplifies to

y + x(dy/dx) = 0

Solving for dy/dx, we have

dy/dx = -y/x

Plugging in the coördinates x=-2, y=6, we have

dy/dx = -y/x = -6/-2 = 3 is the dy/dx at the point (-2, 6).

Answer 2

Find the slope of the tangent line to the graph of the equation xy+12=0 at (-2, 6)

I got m = -3

Yep, you got it.

For others who need help, the tangent to a curve IS the slope, and is the same as the first derivative. So

xy + 12 = 0

xy = -12

y = -12 / x = -12 * x^-1

dy/dx = -12 * (-1 * x^-2) = 12 / x²

At x = -2 we get

dy/dx = 12 / (-2)² = 12 / 4 = 3