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# Find the First Partial Derivatives of the Function. u = 2xy/z

We will be using the concept of differentiation to solve this.

### Answer: The First Partial Derivatives of the Function u = 2xy/z are u_{x} = 2y/z, u_{y} = 2x/z, and u_{z} = -2xy/z^{2}.

Let's solve this step by step.

**Explanation: **

u = 2xy/z

We have to find the values of u_{x}, u_{y}, u_{z. }For this, we have to do the partial differentiation with respect to x, y, and z respectively.

Let's find u_{x} first, here y and z are constants.

u_{x} = du/dx = 2y/z

Let's find u_{y} now, here x and z are constants.

u_{y} = du/dy = 2x/z

Let's find u_{z} first, here x and y are constants.

u_{z} = du/dz = -2xy/z^{2}

### Hence, the first partial derivatives of the function u = 2xy/z are u_{x} = 2y/z, u_{y} = 2x/z, and u_{z} = -2xy/z^{2}.

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