What is (\partial)/(\partial y)(xcos(y)-ye^x) ?

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

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(\partial)/(\partial y)(xcos(y)-ye^x)

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

\int_{1}^{2} 4 x^{2} ln (x) d x

f (x) = \frac{x}{x ^{2} + 4}, at x = 0

\int_{0}^{1} \frac{28}{x ^{2} - 49} d x

6 y^{^{''}} - 6 y^{^{'}} + 15 y = 0

\frac{d}{d x} (xL (x) - x)

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