Riley Barton

2023-03-24

How to implicitly differentiate $\sqrt{xy}=x-2y$?

dalematealreypq3a

Beginner2023-03-25Added 10 answers

First, write it as ${\left(xy\right)}^{\frac{1}{2}}=x-2y$ or ${x}^{\frac{1}{2}}{y}^{\frac{1}{2}}=x-2y$.

Next, differentiate both sides with respect to $x$, assuming that $y$ is a function of $x$. You'll need the Product Rule and the Chain Rule:

$\frac{1}{2}{x}^{-\frac{1}{2}}{y}^{\frac{1}{2}}+\frac{1}{2}{x}^{\frac{1}{2}}{y}^{-\frac{1}{2}}\cdot \frac{dy}{dx}=1-2\frac{dy}{dx}$.

Finalize this equation to obtain $\frac{dy}{dx}$:

$\frac{dy}{dx}(\frac{1}{2}{x}^{\frac{1}{2}}{y}^{-\frac{1}{2}}+2)=1-\frac{1}{2}{x}^{-\frac{1}{2}}{y}^{\frac{1}{2}}$

$\frac{dy}{dx}=\frac{1-\frac{1}{2}{x}^{-\frac{1}{2}}{y}^{\frac{1}{2}}}{\frac{1}{2}{x}^{\frac{1}{2}}{y}^{-\frac{1}{2}}+2}=\frac{1-\frac{\sqrt{y}}{2\sqrt{x}}}{\frac{\sqrt{x}}{2\sqrt{y}}+2}=\frac{2\sqrt{xy}-y}{x+4\sqrt{xy}}$

Next, differentiate both sides with respect to $x$, assuming that $y$ is a function of $x$. You'll need the Product Rule and the Chain Rule:

$\frac{1}{2}{x}^{-\frac{1}{2}}{y}^{\frac{1}{2}}+\frac{1}{2}{x}^{\frac{1}{2}}{y}^{-\frac{1}{2}}\cdot \frac{dy}{dx}=1-2\frac{dy}{dx}$.

Finalize this equation to obtain $\frac{dy}{dx}$:

$\frac{dy}{dx}(\frac{1}{2}{x}^{\frac{1}{2}}{y}^{-\frac{1}{2}}+2)=1-\frac{1}{2}{x}^{-\frac{1}{2}}{y}^{\frac{1}{2}}$

$\frac{dy}{dx}=\frac{1-\frac{1}{2}{x}^{-\frac{1}{2}}{y}^{\frac{1}{2}}}{\frac{1}{2}{x}^{\frac{1}{2}}{y}^{-\frac{1}{2}}+2}=\frac{1-\frac{\sqrt{y}}{2\sqrt{x}}}{\frac{\sqrt{x}}{2\sqrt{y}}+2}=\frac{2\sqrt{xy}-y}{x+4\sqrt{xy}}$

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