What is (\partial)/(\partial y)(e^{-3x}cos(6y)) ?

The answer to (\partial)/(\partial y)(e^{-3x}cos(6y)) is -6e^{-3x}sin(6y)

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(\partial)/(\partial y)(e^{-3x}cos(6y))

\frac{\partial}{\partial y} (e^{- 3 x} cos (6 y))

- 6 e^{- 3 x} sin (6 y)

Solution steps

\frac{\partial}{\partial y} (e^{- 3 x} cos (6 y))

Treat

x

as a constant

Take the constant out:

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

= e^{- 3 x} \frac{\partial}{\partial y} (cos (6 y))

Apply the chain rule:

- sin (6 y) \frac{\partial}{\partial y} (6 y)

= - sin (6 y) \frac{\partial}{\partial y} (6 y)

\frac{\partial}{\partial y} (6 y) = 6

= e^{- 3 x} (- sin (6 y) \cdot 6)

Simplify

= - 6 e^{- 3 x} sin (6 y)

\frac{d y}{d x} = y tan (x)

\int x + 12 x^{2} d x

f (x) = (x - 2)^{2} (x - 3)^{3}

\int \frac{x + 2}{x ^{2} + x} d x

\frac{d y}{d x}, x y = x^{7} y + 72

What is (\partial)/(\partial y)(e^{-3x}cos(6y)) ?
The answer to (\partial)/(\partial y)(e^{-3x}cos(6y)) is -6e^{-3x}sin(6y)