What is implicit (dy)/(dx),(e^{xy})=x+y ?

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

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implicit (dy)/(dx),(e^{xy})=x+y

implicit derivative

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

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

Solution steps

e^{x y} = x + y

Treat

y

y (x)

Differentiate both sides:

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

Isolate

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

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

\int \frac{1}{2 - sin ( x )} d x

n = 1 \sum \infty \frac{1}{1 + 2 + n}

\int \frac{( x - 3 )}{x + 3} d x

x^{2} - 11, - 2

\frac{d}{d x} (\frac{x}{4})

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