What is (\partial)/(\partial x)(e^x+ycos(xy)) ?

The answer to (\partial)/(\partial x)(e^x+ycos(xy)) is e^x-y^2sin(yx)

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(\partial)/(\partial x)(e^x+ycos(xy))

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

e^{x} - y^{2} sin (y x)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

\int (- \frac{1}{x}) d x

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

\frac{d}{d x} (sin (- 2 ln (1 - 2 x)))

\frac{x}{x ^{2} - 3}

y^{^{'}} + \frac{1}{x} y = 4 (x + 7)^{2}

What is (\partial)/(\partial x)(e^x+ycos(xy)) ?
The answer to (\partial)/(\partial x)(e^x+ycos(xy)) is e^x-y^2sin(yx)