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)

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(\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)

\frac{r}{r ^{2} + 8}

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = a, y^{^{'}} (0) = 5

\frac{4}{2 - 4} + \frac{27}{4 + 2}

\int 7 e^{3 x + e^{3 x}} d x

\frac{\partial}{\partial y} (\frac{( x + y )}{( x - y )})

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)