Find the first partial derivatives of the function. u=2xyz ∂u∂x= ∂u∂y= ∂u∂z=

Given: u=2xyz ∂u∂x=2⋅yzxyz−1 ∂u∂x=2yzxyx−1 ∂u∂y=2xyz⋅log⁡x⋅1z ∂u∂y=22⋅log⁡x⋅xyz ∂u∂z=−2xyz⋅log⁡x⋅yz2 ∂u∂z=−2yz2xyx⋅log⁡x Step 2 This is the solution of given question.

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Jerold

Answered question

2021-09-18

Find the first partial derivatives of the function.

u = 2 x^{\frac{y}{z}}

\frac{\partial u}{\partial x} =

\frac{\partial u}{\partial y} =

\frac{\partial u}{\partial z} =

Answer & Explanation

d2saint0

Skilled2021-09-19Added 89 answers

Given:

u = 2 x^{\frac{y}{z}}

\frac{\partial u}{\partial x} = 2 \cdot \frac{y}{z} x^{\frac{y}{z} - 1}

\frac{\partial u}{\partial x} = \frac{2 y}{z} x^{\frac{y}{x} - 1}

\frac{\partial u}{\partial y} = 2 x^{\frac{y}{z}} \cdot \log x \cdot \frac{1}{z}

\frac{\partial u}{\partial y} = \frac{2}{2} \cdot \log x \cdot x^{\frac{y}{z}}

\frac{\partial u}{\partial z} = - 2 x^{\frac{y}{z}} \cdot \log x \cdot \frac{y}{z^{2}}

\frac{\partial u}{\partial z} = - \frac{2 y}{z^{2}} x^{\frac{y}{x}} \cdot \log x

Step 2
This is the solution of given question.

Jeffrey Jordon

Expert2022-03-13Added 2605 answers

Answer is given below (on video)

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