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

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

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

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

e^{x - y}

Solution steps

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

Treat

x, y

as constants

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial z} (z) = 1

= e^{x - y} (0 + 1)

Simplify

= e^{x - y}

\int \frac{1}{9 e ^{- 2 x} + e ^{2 x}} d x

\frac{d}{d x} (ln (sinh (4 x)))

y^{^{'}} + x^{- 1} y = \frac{2}{3} x^{4} y^{4}

y^{^{''}} + 2 y^{^{'}} + y = 12 e^{- t}

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

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