What is (\partial)/(\partial y)(e^x+xcos(y)) ?

The answer to (\partial)/(\partial y)(e^x+xcos(y)) is -xsin(y)

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(\partial)/(\partial y)(e^x+xcos(y))

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

- x sin (y)

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (e^{x}) = 0

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

= 0 - x sin (y)

Simplify

= - x sin (y)

25 y^{^{''}} - 4 y = 0

e^{3 k}

\frac{d}{d x} ((3 x^{4} + 4 x)^{2} (5 x + 8)^{4})

t \to 0 lim (\frac{e ^{6 t} - 1}{t})

y = - 4 e^{- 4 x}

What is (\partial)/(\partial y)(e^x+xcos(y)) ?
The answer to (\partial)/(\partial y)(e^x+xcos(y)) is -xsin(y)