What is implicit ln(xy)=e^{(x+y)} ?

The answer to implicit ln(xy)=e^{(x+y)} is (dy)/(dx)=(e^{x+y}xy-y)/(x(1-e^{y+x)y)}

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implicit ln(xy)=e^{(x+y)}

implicit derivative

ln (x y) = e^{(x + y)}

\frac{d y}{d x} = \frac{e ^{x + y} x y - y}{x ( 1 - e ^{y + x} y )}

Solution steps

ln (x y) = e^{x + y}

Treat

y

y (x)

Differentiate both sides:

\frac{y + x \frac{d y}{d x}}{x y} = e^{x + y} (1 + \frac{d y}{d x})

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{e ^{x + y} x y - y}{x ( 1 - e ^{y + x} y )}

\frac{d y}{d x} = \frac{e ^{x + y} x y - y}{x ( 1 - e ^{y + x} y )}

\frac{d}{d x} (- cos (2 x) 0.2)

\int - 11 cos (ln (x)) d x

f (x) = \frac{x}{x + 1}, at x = 0

f (x) = (2 x - 1) (6 x - 6)^{- 1}

f (x) = \frac{x ^{2}}{x + 2}

What is implicit ln(xy)=e^{(x+y)} ?
The answer to implicit ln(xy)=e^{(x+y)} is (dy)/(dx)=(e^{x+y}xy-y)/(x(1-e^{y+x)y)}