What is (\partial)/(\partial x)(-3x-1e^xcos(y)) ?

The answer to (\partial)/(\partial x)(-3x-1e^xcos(y)) is -3-cos(y)e^x

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(\partial)/(\partial x)(-3x-1e^xcos(y))

\frac{\partial}{\partial x} (- 3 x - 1 e^{x} cos (y))

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

Solution steps

\frac{\partial}{\partial x} (- 3 x - 1 \cdot e^{x} cos (y))

Simplify

- 3 x - 1 \cdot e^{x} cos (y) : - 3 x - e^{x} cos (y)

= \frac{\partial}{\partial x} (- 3 x - e^{x} cos (y))

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial x} (3 x) - \frac{\partial}{\partial x} (e^{x} cos (y))

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

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

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

\int x e^{- x^{2}} d x

\int \frac{y + 1}{y} d y

f (x) = \frac{1}{x + 2}

\frac{\partial}{\partial x} (\frac{x}{3 + y})

\int \frac{1}{x ^{2} + 5 x + 4} d x

What is (\partial)/(\partial x)(-3x-1e^xcos(y)) ?
The answer to (\partial)/(\partial x)(-3x-1e^xcos(y)) is -3-cos(y)e^x