What is (\partial)/(\partial x)(e^{2x}-xcos(xy)) ?

The answer to (\partial)/(\partial x)(e^{2x}-xcos(xy)) is 2e^{2x}-cos(xy)+xysin(xy)

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(\partial)/(\partial x)(e^{2x}-xcos(xy))

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (e^{2 x}) = e^{2 x} \cdot 2

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

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

Simplify

e^{2 x} \cdot 2 - (cos (y x) - y x sin (y x)) : 2 e^{2 x} - cos (x y) + x y sin (x y)

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

\frac{d}{d x} (4 x^{2} - 2 x)

\int \frac{1}{x ^{3} - x} d x

2 y^{^{''}} + y^{^{'}} + 3 y = 3 cos (3 t) + 2 sin (3 t)

\int_{0}^{2} π (2 x - x^{2})^{2} d x

\frac{d}{d x} (e^{x} cos (6 x))

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