What is (\partial)/(\partial y)(-2+e^{-x}cos(y)) ?

The answer to (\partial)/(\partial y)(-2+e^{-x}cos(y)) is -e^{-x}sin(y)

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(\partial)/(\partial y)(-2+e^{-x}cos(y))

\frac{\partial}{\partial y} (- 2 + e^{- x} cos (y))

- e^{- x} sin (y)

Solution steps

\frac{\partial}{\partial y} (- 2 + e^{- x} cos (y))

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{\partial}{\partial y} (2) + \frac{\partial}{\partial y} (e^{- x} cos (y))

\frac{\partial}{\partial y} (2) = 0

\frac{\partial}{\partial y} (e^{- x} cos (y)) = - e^{- x} sin (y)

= - 0 - e^{- x} sin (y)

Simplify

= - e^{- x} sin (y)

\frac{2 x}{( x ^{2} + 1 )}

x \to - 1 lim (2 x^{2} + 3 x + 1)

\int 21 y d y

y^{^{''}} - \frac{2}{r} y^{^{'}} + \frac{2}{r ^{2}} y = 0, y (1) = 1, y (2) = 0

\int 12 x^{2} - 48 x + 36 d x

What is (\partial)/(\partial y)(-2+e^{-x}cos(y)) ?
The answer to (\partial)/(\partial y)(-2+e^{-x}cos(y)) is -e^{-x}sin(y)