What is implicit (dy)/(dx),1+ln(3xy)=e^{3x-y} ?

The answer to implicit (dy)/(dx),1+ln(3xy)=e^{3x-y} is (dy)/(dx)=(3e^{3x-y}xy-y)/(x(1+e^{-y+3x)y)}

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implicit (dy)/(dx),1+ln(3xy)=e^{3x-y}

implicit derivative

\frac{d y}{d x}, 1 + ln (3 x y) = e^{3 x - y}

\frac{d y}{d x} = \frac{3 e ^{3 x - y} x y - y}{x ( 1 + e ^{- y + 3 x} y )}

Solution steps

1 + ln (3 x y) = e^{3 x - y}

Treat

y

y (x)

Differentiate both sides:

\frac{x \frac{d y}{d x} + y}{x y} = e^{3 x - y} (3 - \frac{d y}{d x})

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{3 e ^{3 x - y} x y - y}{x ( 1 + e ^{- y + 3 x} y )}

\frac{d y}{d x} = \frac{3 e ^{3 x - y} x y - y}{x ( 1 + e ^{- y + 3 x} y )}

\frac{\partial}{\partial x} (3 y^{2} - 6 y)

\int (x^{2} + 3)^{5} x d x

n = 1 \sum \infty ((- 1)^{n + 1}) (\frac{cos ( nπ )}{n ^{2}})

6 e^{5 t} cos (2 t)

f (x) = x^{\frac{5}{2}} e^{x}

What is implicit (dy)/(dx),1+ln(3xy)=e^{3x-y} ?
The answer to implicit (dy)/(dx),1+ln(3xy)=e^{3x-y} is (dy)/(dx)=(3e^{3x-y}xy-y)/(x(1+e^{-y+3x)y)}