What is (\partial)/(\partial x)(2e^ysin(-3x)-4y+xy) ?

The answer to (\partial)/(\partial x)(2e^ysin(-3x)-4y+xy) is y-6e^ycos(3x)

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(\partial)/(\partial x)(2e^ysin(-3x)-4y+xy)

\frac{\partial}{\partial x} (2 e^{y} sin (- 3 x) - 4 y + x y)

y - 6 e^{y} cos (3 x)

Solution steps

\frac{\partial}{\partial x} (2 e^{y} sin (- 3 x) - 4 y + x y)

Simplify

2 e^{y} sin (- 3 x) - 4 y + x y : - 4 y + y x - 2 e^{y} sin (3 x)

= \frac{\partial}{\partial x} (- 4 y + y x - 2 e^{y} sin (3 x))

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial x} (4 y) + \frac{\partial}{\partial x} (y x) - \frac{\partial}{\partial x} (2 e^{y} sin (3 x))

\frac{\partial}{\partial x} (4 y) = 0

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

\frac{\partial}{\partial x} (2 e^{y} sin (3 x)) = 6 e^{y} cos (3 x)

= - 0 + y - 6 e^{y} cos (3 x)

Simplify

= y - 6 e^{y} cos (3 x)

7 (ln (y)) y^{^{'}} - x^{3} y = 0

\frac{e ^{- 2 s}}{s ( s + 1.1 )}

\int_{0}^{1} x sin (π x) d x

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

(e^{t^{2}})^{^{'}}

What is (\partial)/(\partial x)(2e^ysin(-3x)-4y+xy) ?
The answer to (\partial)/(\partial x)(2e^ysin(-3x)-4y+xy) is y-6e^ycos(3x)