How do you differentiate $e^x+e^y=e^(x+y)$ ?

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Answer 1 · 2016-06-08T09:59:30

$(dy)/(dx)=(e^x(e^y-1))/(e^y(1-e^x))$

Differentiating $e^x+e^y=e^(x+y)$

$e^x+e^y(dy)/(dx)=e^(x+y)(1+(dy)/(dx))$

or $e^x+e^y(dy)/(dx)=e^(x+y)+e^(x+y)(dy)/(dx)$

or $e^y(dy)/(dx)-e^(x+y)(dy)/(dx)=e^(x+y)-e^x$

or $(e^y-e^(x+y))(dy)/(dx)=(e^(x+y)-e^x)$

or $(dy)/(dx)=(e^(x+y)-e^x)/(e^y-e^(x+y))=(e^x(e^y-1))/(e^y(1-e^x))$

Answer 2 · 2018-07-13T19:28:56

$dy/dx=-e^(y-x)$ .

$e^x+e^y=e^(x+y)=e^x*e^y$ .

$:. (e^x+e^y)/(e^x*e^y)=1$ .

$:. e^x/(e^x*e^y)+e^y/(e^x*e^y)=1$ .

$:. e^-y+e^-x=1$ .

Diff.ing w.r.t. $x$ , we get,

$d/dx(e^-y)+d/dx(e^-x)=d/dx(1)$ .

$:. d/dy(e^-y)*d/dx(-y)+e^-x*d/dx(-x)=0......[because," the Chain Rule]"$ .

$:. e^-y(-dy/dx)+e^-x*-1=0$ .

$:. -dy/dx=e^-x/e^-y=e^(y-x)$ .

$rArr dy/dx=-e^(y-x)$ , as Respected Shwetank Mauria has

readily derived!

How do you differentiate e^x+e^y=e^(x+y)e^x+e^y=e^(x+y)?