# How do you use implicit differentiation to find the slope of the curve given 1/x^3+1/y^3=2 at (2,2)?

Nov 15, 2016

This is either a 'trick' question or there is an error. The point $\left(2 , 2\right)$ is not on the graph of $\frac{1}{x} ^ 3 + \frac{1}{y} ^ 3 = 2$

#### Explanation:

We can find $\frac{\mathrm{dy}}{\mathrm{dx}}$ using implicit differentiation.

${x}^{-} 3 + {y}^{-} 3 = 2$ (or some other constant. If we use $\frac{1}{4}$ then $\left(2 , 2\right)$ is on the graph.)

Differentiate both sides with respect to $x$:

$- 3 {x}^{-} 4 - 3 {y}^{-} 4 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$- \frac{3}{x} ^ 4 - \frac{3}{y} ^ 4 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{4} / {x}^{4}$

At any point $\left(a , a\right)$ that is on the graph, the slope of the tangent line is $- 1$.