How do you implicitly differentiate $ln (x y) = x + y$ $ln(xy)=x+y$ ?

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Answer 1 · 2016-07-01T11:43:23

$\frac{x y - y}{x - x y}$ $(xy-y)/(x-xy)$

Given expression

$ln (x y) = x + y$ $ln(xy)=x+y$

$\Rightarrow ln x + ln y = x + y$ $=>lnx+lny=x+y$

Differentiating w.r. to x we can write

$\frac{d (ln x)}{d x} + \frac{d (ln y)}{d x} = \frac{d (x)}{d x} + \frac{d (y)}{d x}$ $(d(lnx))/(dx)+(d(lny))/(dx)=(d(x))/(dx)+(d(y))/(dx)$

$\Rightarrow \frac{1}{x} + \frac{1}{y} \cdot \frac{d y}{d x} = 1 + \frac{d y}{d x}$ $=>1/x+1/y*(dy)/(dx)=1+(dy)/(dx)$

$\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x} - \frac{d y}{d x} = 1 - \frac{1}{x}$ $=>1/y*(dy)/(dx)-(dy)/(dx)=1-1/x$

$\Rightarrow (\frac{1}{y} - 1) \frac{d y}{d x} = \frac{x - 1}{x}$ $=>(1/y-1)(dy)/(dx)=(x-1)/x$

$\Rightarrow (\frac{1 - y}{y}) \frac{d y}{d x} = \frac{x - 1}{x}$ $=>((1-y)/y)(dy)/(dx)=(x-1)/x$

$\Rightarrow \frac{d y}{d x} = \frac{x - 1}{x} \times \frac{y}{1 - y} = \frac{x y - y}{x - x y}$ $=>(dy)/(dx)=(x-1)/x xx(y)/(1-y)=(xy-y)/(x-xy)$

How do you implicitly differentiate ln(xy)=x+y ln(xy)=x+yln(xy)=x+y ?