What is implicit (dy)/(dx),y^2cos(1/y)=2x+2y ?

The answer to implicit (dy)/(dx),y^2cos(1/y)=2x+2y is (dy)/(dx)= 2/(2ycos(1/y)+sin(1/y)-2)

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implicit (dy)/(dx),y^2cos(1/y)=2x+2y

implicit derivative

\frac{d y}{d x}, y^{2} cos (\frac{1}{y}) = 2 x + 2 y

\frac{d y}{d x} = \frac{2}{2 y cos ( \frac{1}{y} ) + sin ( \frac{1}{y} ) - 2}

Solution steps

y^{2} cos (\frac{1}{y}) = 2 x + 2 y

Treat

y

y (x)

Differentiate both sides:

2 cos (\frac{1}{y}) y \frac{d y}{d x} + sin (\frac{1}{y}) \frac{d y}{d x} = 2 + 2 \frac{d y}{d x}

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{2}{2 y cos ( \frac{1}{y} ) + sin ( \frac{1}{y} ) - 2}

\frac{d y}{d x} = \frac{2}{2 y cos ( \frac{1}{y} ) + sin ( \frac{1}{y} ) - 2}

\int \frac{4 e ^{2 x}}{e ^{2 x} + 14 e ^{x} + 45} d x

\frac{d}{d x} ((\frac{5}{x} + x) (\frac{5}{x} - x))

y^{^{'}} = (x - 2) e^{- 2 y}

\int \frac{1}{9 e ^{- 6 x} + e ^{6 x}} d x

3 y^{^{''}} + 27 y = 0

What is implicit (dy)/(dx),y^2cos(1/y)=2x+2y ?
The answer to implicit (dy)/(dx),y^2cos(1/y)=2x+2y is (dy)/(dx)= 2/(2ycos(1/y)+sin(1/y)-2)