What is (\partial)/(\partial z)(xe^z+xy+z^3-4) ?

The answer to (\partial)/(\partial z)(xe^z+xy+z^3-4) is e^zx+3z^2

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(\partial)/(\partial z)(xe^z+xy+z^3-4)

\frac{\partial}{\partial z} (x e^{z} + x y + z^{3} - 4)

e^{z} x + 3 z^{2}

Solution steps

\frac{\partial}{\partial z} (x e^{z} + x y + z^{3} - 4)

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial z} (x e^{z}) + \frac{\partial}{\partial z} (x y) + \frac{\partial}{\partial z} (z^{3}) - \frac{\partial}{\partial z} (4)

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

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

\frac{\partial}{\partial z} (z^{3}) = 3 z^{2}

\frac{\partial}{\partial z} (4) = 0

= x e^{z} + 0 + 3 z^{2} - 0

Simplify

= e^{z} x + 3 z^{2}

\int e^{x} 1 - e^{2 x} d x

e^{1 - x}

x \frac{d y}{d x} + y + y^{2} = 0

\frac{( 1 + tan ( x ) ) ^{2}}{sec ( x )}

\int_{1}^{2} (4 x - \frac{2}{x ^{2}}) d x

What is (\partial)/(\partial z)(xe^z+xy+z^3-4) ?
The answer to (\partial)/(\partial z)(xe^z+xy+z^3-4) is e^zx+3z^2