What is (\partial)/(\partial x)(2xyz-ye^{2xyz}) ?

The answer to (\partial)/(\partial x)(2xyz-ye^{2xyz}) is 2yz-2y^2ze^{2yzx}

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(\partial)/(\partial x)(2xyz-ye^{2xyz})

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

2 yz - 2 y^{2} z e^{2 yz x}

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (2 x yz) = 2 yz

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

= 2 yz - 2 y^{2} z e^{2 yz x}

(3.1) (1.7)

n = 1 \sum \infty (- 3)^{n} x^{n}

\frac{d}{d x} (- sin (x^{2}) 2 x)

\frac{d y}{d x} + 9 (tan (9 x)) y = cos (9 x)

x \to 4 - lim (\frac{∣ 2 x ∣}{x})

What is (\partial)/(\partial x)(2xyz-ye^{2xyz}) ?
The answer to (\partial)/(\partial x)(2xyz-ye^{2xyz}) is 2yz-2y^2ze^{2yzx}