What is (\partial)/(\partial x)(5+e^{-xy}cos(y)) ?

The answer to (\partial)/(\partial x)(5+e^{-xy}cos(y)) is -e^{-xy}ycos(y)

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(\partial)/(\partial x)(5+e^{-xy}cos(y))

\frac{\partial}{\partial x} (5 + e^{- x y} cos (y))

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

Solution steps

\frac{\partial}{\partial x} (5 + 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} (5) + \frac{\partial}{\partial x} (e^{- x y} cos (y))

\frac{\partial}{\partial x} (5) = 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)

x \to 7 lim (\frac{x ^{2} - 5 x - 14}{x - 7})

\int_{1}^{e} \frac{x ^{2} - 1}{x} d x

y^{^{''}} - 6 y^{^{'}} + 6 y = 0

\frac{\partial}{\partial x} (sin (x^{2} y))

\frac{d}{d x} (sin ((π x)^{5}))

What is (\partial)/(\partial x)(5+e^{-xy}cos(y)) ?
The answer to (\partial)/(\partial x)(5+e^{-xy}cos(y)) is -e^{-xy}ycos(y)