What is (\partial)/(\partial x)(e^{-x}cos(-8y)) ?

The answer to (\partial)/(\partial x)(e^{-x}cos(-8y)) is -e^{-x}cos(8y)

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(\partial)/(\partial x)(e^{-x}cos(-8y))

\frac{\partial}{\partial x} (e^{- x} cos (- 8 y))

- e^{- x} cos (8 y)

Solution steps

\frac{\partial}{\partial x} (e^{- x} cos (- 8 y))

Simplify

e^{- x} cos (- 8 y) : e^{- x} cos (8 y)

= \frac{\partial}{\partial x} (e^{- x} cos (8 y))

Treat

y

as a constant

Take the constant out:

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

= cos (8 y) \frac{\partial}{\partial x} (e^{- x})

Apply the chain rule:

e^{- x} \frac{\partial}{\partial x} (- x)

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

\frac{\partial}{\partial x} (- x) = - 1

= cos (8 y) e^{- x} (- 1)

Simplify

= - e^{- x} cos (8 y)

f (x) = \frac{12}{ln ( x )}

\frac{1}{( s ^{2} + 3 s + 2 )}

x \to \infty lim (cos (x))

\int_{- 2}^{3} \frac{20}{x ^{4}} d x

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

What is (\partial)/(\partial x)(e^{-x}cos(-8y)) ?
The answer to (\partial)/(\partial x)(e^{-x}cos(-8y)) is -e^{-x}cos(8y)