rancuri5a

2022-09-06

How do you find the derivative of y in the equation $\mathrm{ln}\left(xy\right)=x+y$?

Marshall Horne

Beginner2022-09-07Added 8 answers

You have to remember that in this "implicit" function y is a function of x so deriving you get: $\frac{1}{xy}\cdot (y+x\frac{dy}{dx})=1+\frac{dy}{dx}$ where you used the Chain Rule on the ln

Then:

$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{dx}$

So you collect $\frac{dy}{dx}$ and get:

$\frac{dy}{dx}=\frac{1-\frac{1}{x}}{\frac{1}{y}-1}=\left(\frac{x-1}{x}\right)\left(\frac{y}{1-y}\right)=\frac{y(x-1)}{x(1-y)}$

Then:

$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{dx}$

So you collect $\frac{dy}{dx}$ and get:

$\frac{dy}{dx}=\frac{1-\frac{1}{x}}{\frac{1}{y}-1}=\left(\frac{x-1}{x}\right)\left(\frac{y}{1-y}\right)=\frac{y(x-1)}{x(1-y)}$

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