What is (\partial)/(\partial y)(e^xcos(x+y)) ?

The answer to (\partial)/(\partial y)(e^xcos(x+y)) is -e^xsin(x+y)

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(\partial)/(\partial y)(e^xcos(x+y))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

- sin (x + y) \frac{\partial}{\partial y} (x + y)

= - sin (x + y) \frac{\partial}{\partial y} (x + y)

\frac{\partial}{\partial y} (x + y) = 1

= e^{x} (- sin (x + y) \cdot 1)

Simplify

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

\frac{\partial}{\partial x} (\frac{2}{3} (1 - x)^{\frac{3}{2}})

n = 0 \sum \infty (\frac{1}{9})^{n}

\frac{d}{d x} (x + \frac{3}{x} + 5)

\frac{\partial}{\partial x} (tan (\frac{x}{2}))

What is (\partial)/(\partial y)(e^xcos(x+y)) ?
The answer to (\partial)/(\partial y)(e^xcos(x+y)) is -e^xsin(x+y)