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

The answer to (\partial)/(\partial x)((e^{2y}-ycos(xy))dx) is d(y^2xsin(yx)+e^{2y}-ycos(yx))

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(\partial)/(\partial x)((e^{2y}-ycos(xy))dx)

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

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

Solution steps

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

Treat

y, d

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{2 y} - y cos (x y), g = x

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

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

\frac{\partial}{\partial x} (x) = 1

= d (y^{2} sin (y x) x + 1 \cdot (e^{2 y} - y cos (x y)))

Simplify

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

\int e^{x} 1 + e^{2 x} d x

y = x^{4^{x}}

\int 9 x e^{5 x^{2}} d x

\frac{d}{d x} (A (e^{x} + e^{x} x))

x^{2} + x y + 2 y^{2} = 4, (- 2, 1)

What is (\partial)/(\partial x)((e^{2y}-ycos(xy))dx) ?
The answer to (\partial)/(\partial x)((e^{2y}-ycos(xy))dx) is d(y^2xsin(yx)+e^{2y}-ycos(yx))