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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

f (x) = ln (1 - x)

y^{^{'}} = - k y^{\frac{1}{2}}

\frac{d}{d x} (\frac{4}{3} x^{\frac{3}{2}})

\int csc^{4} (x) d x

cos^{l n (7 x)} (x)

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