What is (\partial)/(\partial x)(-e^ysin(x)+4y+7yx) ?

The answer to (\partial)/(\partial x)(-e^ysin(x)+4y+7yx) is 7y-e^ycos(x)

(\partial)/(\partial x)(-e^ysin(x)+4y+7yx)

\frac{\partial}{\partial x} (- e^{y} sin (x) + 4 y + 7 y x)

7 y - e^{y} cos (x)

Solution steps

\frac{\partial}{\partial x} (- e^{y} sin (x) + 4 y + 7 y x)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial x} (e^{y} sin (x)) + \frac{\partial}{\partial x} (4 y) + \frac{\partial}{\partial x} (7 y x)

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

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

\frac{\partial}{\partial x} (7 y x) = 7 y

= - e^{y} cos (x) + 0 + 7 y

Simplify

= 7 y - e^{y} cos (x)

x \to 3 lim (\frac{5}{( x - 3 ) ^{2}})

x \to 2 lim (3 x^{2} - 5 x + 2)

e^{y} d x - e^{- x} d y = 0, y (0) = 0

d er i v a t i v e f (x) = \frac{x + 6}{e ^{x}}

\frac{d}{d x} (x cos (y) - y cos (x))

What is (\partial)/(\partial x)(-e^ysin(x)+4y+7yx) ?
The answer to (\partial)/(\partial x)(-e^ysin(x)+4y+7yx) is 7y-e^ycos(x)