What is (\partial)/(\partial x)(e^{-3y}cos(-3x)) ?

The answer to (\partial)/(\partial x)(e^{-3y}cos(-3x)) is -3e^{-3y}sin(3x)

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

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

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

Solution steps

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

Simplify

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

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

- sin (3 x) \frac{\partial}{\partial x} (3 x)

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

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

= e^{- 3 y} (- sin (3 x) \cdot 3)

Simplify

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

\int \frac{1}{1 + t ^{2}} d t

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x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = x^{\frac{5}{2}}

\frac{\partial}{\partial x} (7 ln (\frac{3 x y}{7 z}))

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What is (\partial)/(\partial x)(e^{-3y}cos(-3x)) ?
The answer to (\partial)/(\partial x)(e^{-3y}cos(-3x)) is -3e^{-3y}sin(3x)