# How do you use implicit differentiation to find dy/dx given 2^sqrt(xy)=x?

Jan 17, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln \frac{x}{x \ln 2} ^ 2 \left(2 - \ln x\right)$

#### Explanation:

Differentiate both members of the equation with respect to $x$ keeping in mind that:

$\frac{d}{\mathrm{dx}} f \left(y \left(x\right)\right) = \frac{\mathrm{df}}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} {2}^{\sqrt{x y}} = \frac{d}{\mathrm{dx}} x$

$\ln 2 \left(\frac{y + x y '}{2 \sqrt{x y}}\right) {2}^{\sqrt{x y}} = 1$

Substitute ${2}^{\sqrt{x y}} = x$ from the original equation:

$\ln 2 \left(\frac{y + x y '}{2 \sqrt{x y}}\right) x = 1$

Solving for $y '$:

$\left(y + x y '\right) = \frac{2}{\ln} 2 \frac{\sqrt{x y}}{x}$

$y ' = - \frac{y}{x} + \frac{2}{\ln} 2 \frac{\sqrt{x y}}{x} ^ 2$

Now note that:

${2}^{\sqrt{x y}} = x \implies \sqrt{x y} = \ln \frac{x}{\ln} 2$

and so:

$x y = {\ln}^{2} \frac{x}{\ln} ^ 2 2$

$y = \frac{1}{x} {\ln}^{2} \frac{x}{\ln} ^ 2 2$

Substituting in the expression of $y '$ we have:

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

You can also first make $y$ explicit from the original equation and then differentiate:

$y = \frac{1}{x} {\ln}^{2} \frac{x}{\ln} ^ 2 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(- \frac{1}{x} ^ 2\right) {\ln}^{2} \frac{x}{\ln} ^ 2 2 + \frac{1}{x} \frac{1}{\ln} ^ 2 2 2 \ln \frac{x}{x} = \ln \frac{x}{x \ln 2} ^ 2 \left(2 - \ln x\right)$

Obtaining naturally the same result.

Jan 17, 2017

x and y > 0. $y ' = \frac{1}{x} \left(\frac{2}{\ln} 2 \sqrt{\frac{y}{x}} - y\right)$.

#### Explanation:

graph{x-2^sqrt(xy)=0x^2 [-39.94, 39.94, -19.96, 19.98]} To make the function real, $x > 0$, giving $y > 0$.

$y \to \infty$, as $x \to {0}_{+}$.

Using logarithmic differentiation,

$\ln 2 \sqrt{x y} ' = \left(\ln x\right) '$, giving

$\frac{1}{2} \left(y ' \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}}\right) = \frac{1}{x \ln 2}$. So,

$y ' = \left(\frac{2}{x \ln 2} - \sqrt{\frac{y}{x}}\right) \sqrt{\frac{y}{x}}$

$= \frac{1}{x} \left(\frac{2}{\ln} 2 \sqrt{\frac{y}{x}} - y\right)$