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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (8) = 0

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

= 0 - y e^{- y x} cos (y)

Simplify

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

f (x) = \frac{1}{2 + 3 x}, (2, \frac{1}{8})

y = 1 - 2 x

g (x) = (x^{2} + 1)^{3} (x^{2} + 2)^{6}

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

f (x) = 4 x^{4} - \frac{1}{4 x ^{4}}

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