What is (\partial)/(\partial y)((z)(xe^{yz}+ye^{xz}+3xyz-2)) ?

The answer to (\partial)/(\partial y)((z)(xe^{yz}+ye^{xz}+3xyz-2)) is z(e^{zy}zx+3zx+e^{zx})

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(\partial)/(\partial y)((z)(xe^{yz}+ye^{xz}+3xyz-2))

\frac{\partial}{\partial y} ((z) (x e^{yz} + y e^{x z} + 3 x yz - 2))

z (e^{zy} z x + 3 z x + e^{z x})

Solution steps

\frac{\partial}{\partial y} ((z) (x e^{yz} + y e^{x z} + 3 x yz - 2))

Treat

z, x

as constants

Take the constant out:

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

= z \frac{\partial}{\partial y} (x e^{yz} + y e^{x z} + 3 x yz - 2)

Apply the Sum/Difference Rule:

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

= z (\frac{\partial}{\partial y} (x e^{yz}) + \frac{\partial}{\partial y} (y e^{x z}) + \frac{\partial}{\partial y} (3 x yz) - \frac{\partial}{\partial y} (2))

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

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

\frac{\partial}{\partial y} (3 x yz) = 3 z x

\frac{\partial}{\partial y} (2) = 0

= z (x e^{yz} z + e^{z x} + 3 z x - 0)

Simplify

= z (e^{zy} z x + 3 z x + e^{z x})

\frac{2 s + 3}{( s - 1 ) ^{3} ( s ^{3} + 1 )}

\frac{1}{( a - x ) ( b - x )}

(cos (\frac{θ}{2}) + sin (\frac{θ}{2}) e^{i φ})^{2}

e^{3 x y} \cdot 3 (y + x \frac{d y}{d x})

50 (24 - 23 e^{- \frac{t}{200}})

What is (\partial)/(\partial y)((z)(xe^{yz}+ye^{xz}+3xyz-2)) ?
The answer to (\partial)/(\partial y)((z)(xe^{yz}+ye^{xz}+3xyz-2)) is z(e^{zy}zx+3zx+e^{zx})