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

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

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

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

2 e^{y} cos (x) - 4 y

Solution steps

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

Simplify

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

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (3 y) = 0

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

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

= 0 + 2 e^{y} cos (x) - 4 y

Simplify

= 2 e^{y} cos (x) - 4 y

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