# How do you differentiate sqrt (1-x^4) + sqrt (1-y^4)= k(x^2-y^2)?

Feb 25, 2017

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

#### Explanation:

Writing as:

${\left(1 - {x}^{4}\right)}^{\frac{1}{2}} + {\left(1 - {y}^{4}\right)}^{\frac{1}{2}} = k \left({x}^{2} - {y}^{2}\right)$

When we differentiate this, we will use the chain rule frequently. Recall that, as an example, the derivative with respect to $x$ of ${y}^{3}$ is not $3 {y}^{2}$ but $3 {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$.

Differentiating:

$\frac{1}{2} {\left(1 - {x}^{4}\right)}^{- \frac{1}{2}} \left(- 4 {x}^{3}\right) + \frac{1}{2} {\left(1 - {y}^{4}\right)}^{- \frac{1}{2}} \left(- 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}\right) = k \left(2 x - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Expanding and rewriting:

$\frac{- 2 {x}^{3}}{\sqrt{1 - {x}^{4}}} + \frac{- 2 {y}^{3}}{\sqrt{1 - {y}^{4}}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 k x - 2 k y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Rewriting to solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$, the derivative:

$2 k y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - \frac{2 {y}^{3}}{\sqrt{1 - {y}^{4}}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 k x + \frac{2 {x}^{3}}{\sqrt{1 - {x}^{4}}}$

Factoring $\frac{\mathrm{dy}}{\mathrm{dx}}$ and dividing through by $2$:

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

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

Personally, this is as far as I want to go, but we can "simplify" by getting common denominators:

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

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