# How do you find the implicit differentiation of 1+lnxy=3^(x-y)?

Given:
$1 + \setminus \ln \left(x y\right) = {3}^{x - y} \setminus \ldots \ldots . \left(1\right)$

$1 + \setminus \ln \left(x\right) + \setminus \ln \left(y\right) = {3}^{x - y}$

differentiating w.r.t. $x$ as follows

$\setminus \frac{d}{\mathrm{dx}} \left(1 + \setminus \ln \left(x\right) + \setminus \ln \left(y\right)\right) = \setminus \frac{d}{\mathrm{dx}} \left({3}^{x - y}\right)$

$0 + \setminus \frac{1}{x} + \setminus \frac{1}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = {3}^{x - y} \setminus \ln \left(3\right) \setminus \frac{d}{\mathrm{dx}} \left(x - y\right)$

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

$\setminus \frac{1}{x} + \setminus \frac{1}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 + \setminus \ln \left(x y\right)\right) \setminus \ln \left(3\right) \left(1 - \setminus \frac{\mathrm{dy}}{\mathrm{dx}}\right)$ (from(1))

$\left[\left(1 + \setminus \ln \left(x y\right)\right) \setminus \ln \left(3\right) + \setminus \frac{1}{y}\right] \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 + \setminus \ln \left(x y\right)\right) \setminus \ln \left(3\right) - \setminus \frac{1}{x}$

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{\left(1 + \setminus \ln \left(x y\right)\right) \setminus \ln \left(3\right) - \setminus \frac{1}{x}}{\left(1 + \setminus \ln \left(x y\right)\right) \setminus \ln \left(3\right) + \setminus \frac{1}{y}}$

Jun 24, 2018

dy/dx=((ln(3)*3^(x+y))/(3^(2y))-1/x)/(((ln(3)*3^(x+y))/3^(2y)+1/y)

#### Explanation:

First separate the equation into easier pieces.
$1 + \ln x + \ln y = \frac{{3}^{x}}{{3}^{y}}$

Differentiate left side

$\frac{1}{x} + \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

Differentiate right side using quotient rule

$\frac{\ln \left(3\right) \cdot {3}^{x} \cdot {3}^{y} - \ln \left(3\right) \cdot {3}^{x} \cdot {3}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}}{{\left({3}^{y}\right)}^{2}}$

Set equal again
$\frac{1}{x} + \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\ln \left(3\right) \cdot {3}^{x} \cdot {3}^{y} - \ln \left(3\right) \cdot {3}^{x} \cdot {3}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}}{{\left({3}^{y}\right)}^{2}}$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$

dy/dx=((ln(3)*3^(x+y))/(3^(2y))-1/x)/(((ln(3)*3^(x+y))/3^(2y)+1/y)