What is (\partial)/(\partial x)(cos(y)+ye^x) ?

The answer to (\partial)/(\partial x)(cos(y)+ye^x) is e^xy

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(\partial)/(\partial x)(cos(y)+ye^x)

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

e^{x} y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (cos (y)) = 0

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

= 0 + y e^{x}

Simplify

= e^{x} y

\int \frac{16}{x ^{2} 16 - x ^{2}} d x

y^{^{'}} = - \frac{y}{( x + \frac{2}{y} )}

y^{^{'}} + y = b

x + 2

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

What is (\partial)/(\partial x)(cos(y)+ye^x) ?
The answer to (\partial)/(\partial x)(cos(y)+ye^x) is e^xy