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

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

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

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

y^{2} sin (y x)

Solution steps

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

\frac{\partial}{\partial x} (e^{2 y}) = 0

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

= 0 - (- y^{2} sin (y x))

Simplify

= y^{2} sin (y x)

y = e^{x}, y = e^{- 3 x}, x = ln (3)

\frac{\partial}{\partial x} (x y (x - 2) (y - 3))

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

f (x) = x ln (x)

\int 3 x^{2} + 4 d x

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