What is (\partial)/(\partial x)(2e^{xyz}xz) ?

The answer to (\partial)/(\partial x)(2e^{xyz}xz) is 2z(e^{xyz}xyz+e^{xyz})

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(\partial)/(\partial x)(2e^{xyz}xz)

\frac{\partial}{\partial x} (2 e^{x yz} x z)

2 z (e^{x yz} x yz + e^{x yz})

Solution steps

\frac{\partial}{\partial x} (2 e^{x yz} x z)

Treat

y, z

as constants

Take the constant out:

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

= 2 z \frac{\partial}{\partial x} (e^{x yz} x)

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{x yz}, g = x

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

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

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

= 2 z (e^{x yz} yz x + 1 \cdot e^{x yz})

Simplify

= 2 z (e^{x yz} x yz + e^{x yz})

x^{2} y^{^{'}} = 3 x y + 4 x^{6}

h \to 0 lim (\frac{∣ 3 + h ∣ - ∣ 3 ∣}{h})

\frac{d}{d x} (x - ∣ ln ∣ x ∣ ∣)

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

x^{^{'}} = x^{\frac{4}{5}}

What is (\partial)/(\partial x)(2e^{xyz}xz) ?
The answer to (\partial)/(\partial x)(2e^{xyz}xz) is 2z(e^{xyz}xyz+e^{xyz})