Ln(xy) = e ^(x + y) how would one differentiate this equation?​

Question

`ln(xy) = e ^(x + y)`
how would one differentiate this equation?​

Answer 1

`~~~~~~\ln(xy)= e^(x+y)\\\\\\\implies (d)/(dx) \ln(xy) = (d)/(dx)\left(e^(x+y) \right)\\\\\\\implies \left(\frac 1{xy}\right) (d)/(dx)(xy) =\left( e^(x+y) \right)(d)/(dx) (x+y)~~~~~~~~~~~~~~~~~;[\text{Chain rule}]\\\\\\\implies \left( \frac 1{xy} \right) \left(y + x(dy)/(dx) \right) =\left(e^(x+y)\right) \left( 1 + (dy)/(dx) \right) \\\\\\`

`\implies \frac 1x + \frac 1y\cdot (dy)/(dx) = e^(x+y)+ \left(e^(x+y) \right) (dy)/(dx)\\\\\\\implies y + x (dy)/(dx) = xy e^(x+y)+ \left(xy e^(x+y) \right)(dy)/(dx)~~~~~~~~~;[\text{Multiply both sides by}~ xy]\\\\\\`

`\implies x (dy)/(dx) -\left(xye^(x+y) \right) (dy)/(dx)= xye^(x+y) -y\\\\\\\implies \left(x - x y e^(x+y) \right) (dy)/(dx) = xye^(x+y) - y\\\\\\\implies (dy)/(dx) = (xye^(x+y) - y)/(x - x y e^(x+y))\\\\\\\implies (dy)/(dx) = (y \left(xe^(x+y) -1 \right))/(x \left(1- ye^(x+y) \right))`