What is implicit e^{xy}+y=x-1 ?

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

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implicit e^{xy}+y=x-1

implicit derivative

e^{x y} + y = x - 1

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

Solution steps

e^{x y} + y = x - 1

Treat

y

y (x)

Differentiate both sides:

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

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}

\frac{x}{x ^{2} + 4}

\frac{\partial}{\partial x} (3 e^{x - y})

\int_{6}^{8} \frac{52}{( x - 6 ) ^{3}} d x

\int \frac{18 x}{1 - 9 x ^{2}} d x

\int 3 x (6 - x^{2}) d x

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