# How do you use implicit differentiation to find the slope of the curve given x^(1/4)+y^(1/4)=4 at (16,16)?

Feb 2, 2017

$- 1$.

#### Explanation:

By the power rule:

$\frac{1}{4} {x}^{- \frac{3}{4}} + \frac{1}{4} {y}^{- \frac{3}{4}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$\frac{1}{4} {y}^{- \frac{3}{4}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - \frac{1}{4} {x}^{- \frac{3}{4}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \frac{1}{4} {x}^{- \frac{3}{4}}}{\frac{1}{4} {y}^{- \frac{3}{4}}}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{\frac{3}{4}} / \left({x}^{\frac{3}{4}}\right)$

Now simply evaluate the given point within the derivative.

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{16}^{\frac{3}{4}}}{{16}^{\frac{3}{4}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{8}{8}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 1$

Hopefully this helps!