# What the slope of x^4y^4=16 at (2,1) ?

May 13, 2018

The slope (of the tangent) to the curve at the given coordinate is
$- \frac{1}{2}$

#### Explanation:

We seek the slope (of the tangent) to the curve:

${x}^{4} {y}^{4} = 16$

at the coordinate $\left(2 , 1\right)$.

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So if we differentiate the equation implicitly, and apply the product rule, then we have:

$\left({x}^{4}\right) \left(\frac{d}{\mathrm{dx}} {y}^{4}\right) + \left(\frac{d}{\mathrm{dx}} {x}^{4}\right) \left({y}^{4}\right) = 0$

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

So if the slope of the tangent at the given coordinate $\left(2 , 1\right)$ is $m$, then we have:

$\therefore \left({2}^{4}\right) \left(4 \cdot {1}^{3} \cdot m\right) + \left(4 \cdot {2}^{3}\right) \left({1}^{4}\right) = 0$

$\therefore 16 \left(4 m\right) + 4 \cdot 8 = 0$

$\therefore 2 m + 1 = 0$

$\therefore m = - \frac{1}{2}$