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

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

(\partial)/(\partial y)(e^{-x}sin(x+y))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\frac{d}{d x} (\frac{x ^{2}}{cos ( x )})

\int \frac{1}{2} e^{2 x} cos (x) d x

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

d er i v a t i v e q (t) = sin^{2} (\frac{t ^{4}}{t ^{5} - 6 t})

\int_{1}^{4} 1 t ln (t) d t

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