Resolved: "Compute the first-order partial derivatives. z=x2+y2"

compagnia04

Answered question

2022-01-16

Compute the first-order partial derivatives.

z = x^{2} + y^{2}

amarantha41

Beginner2022-01-17Added 38 answers

Step 1
Explanation:
Given that,

z = x^{2} + y^{2}

Concept used:
Here we treat y as a constant.
Sum and difference rule of derivatives:

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

Step 2
Calculation:

z = x^{2} + y^{2}

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

= \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial x} (y^{2})

=2x+0
=2x
Hence, the first-order partial derivative of

z = x^{2} + y^{2}

is 2x

Carl Swisher

Beginner2022-01-18Added 28 answers

The required is to compute the first-order partial derivatives:

z = x^{2} + y^{2}

The first-order partial derivatives:

∴ z_{x} = \frac{\partial}{\partial x} (x^{2} + y^{2})

=2x+0
=2x

∴ z_{y} = \frac{\partial}{\partial y} (x^{2} + y^{2})

=0+2y
=2y
Result:

z_{x} = 2 x, z_{y} = 2 y

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