# How do you differentiate e^sqrt(xy)-xy=6?

Mar 24, 2016

We need to differentiate
${e}^{\sqrt{x y}} - x y = 6$
We see that there are three terms.
In the first term we use chain rule along-with product rule,
$\frac{d}{\mathrm{dx}} \left(u \cdot v\right) = u \cdot \frac{\mathrm{dv}}{\mathrm{dx}} + v \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
for the second term product rule and
since the third term is a constant it derivative $= 0$

$D \left({e}^{\sqrt{x y}}\right) - D \left(x y\right) = D \left(6\right)$
The exponent of first term can be written as ${\left(x y\right)}^{\frac{1}{2}}$
${e}^{\sqrt{x y}} \cdot D {\left(x y\right)}^{\frac{1}{2}} - D \left(x y\right) = 0$

${e}^{\sqrt{x y}} \cdot \frac{1}{2} \cdot {\left(x y\right)}^{- \frac{1}{2}} \cdot D \left(x y\right) - \left(x \cdot y ' + y\right) = 0$
${e}^{\sqrt{x y}} \cdot \frac{1}{2} \cdot {\left(x y\right)}^{- \frac{1}{2}} \cdot \left(x \cdot y ' + y\right) - \left(x \cdot y ' + y\right) = 0$
$\frac{{e}^{\sqrt{x y}}}{2 \cdot \sqrt{x y}} \cdot \left(x \cdot y ' + y\right) - \left(x \cdot y ' + y\right) = 0$
Let $\frac{{e}^{\sqrt{x y}}}{2 \cdot \sqrt{x y}} = G \left(x , y\right)$
Above expression becomes
$G \cdot \left(x \cdot y ' + y\right) - \left(x \cdot y ' + y\right) = 0$,
$G \cdot x \cdot y ' + G \cdot y - x \cdot y ' - y = 0$, solving for $y '$
$\left(G - 1\right) \cdot x \cdot y ' + \left(G - 1\right) \cdot y = 0$,
dividing both sides with $\left(G - 1\right)$
$x \cdot y ' + y = 0$
$y ' = - \frac{y}{x}$ with the condition $\left(G - 1\right) \ne 0$