What is (\partial)/(\partial x)(-7e^ysin(2x)+y+5xy) ?

The answer to (\partial)/(\partial x)(-7e^ysin(2x)+y+5xy) is 5y-14e^ycos(2x)

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(\partial)/(\partial x)(-7e^ysin(2x)+y+5xy)

\frac{\partial}{\partial x} (- 7 e^{y} sin (2 x) + y + 5 x y)

5 y - 14 e^{y} cos (2 x)

Solution steps

\frac{\partial}{\partial x} (- 7 e^{y} sin (2 x) + y + 5 x y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial x} (7 e^{y} sin (2 x)) + \frac{\partial}{\partial x} (y) + \frac{\partial}{\partial x} (5 x y)

\frac{\partial}{\partial x} (7 e^{y} sin (2 x)) = 14 e^{y} cos (2 x)

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

\frac{\partial}{\partial x} (5 x y) = 5 y

= - 14 e^{y} cos (2 x) + 0 + 5 y

Simplify

= 5 y - 14 e^{y} cos (2 x)

\frac{\partial}{\partial x} (x^{- 7} y^{2} + x y^{2} + 2 x y)

n = 0 \sum \infty (2 n)!

x \to - 2 lim (9 x^{2})

x \to - 3 lim (\frac{x ^{2}}{9 - x ^{2}})

y^{^{'}} + \frac{y}{x} = x y^{2}

What is (\partial)/(\partial x)(-7e^ysin(2x)+y+5xy) ?
The answer to (\partial)/(\partial x)(-7e^ysin(2x)+y+5xy) is 5y-14e^ycos(2x)