Best solutions to - "Find both first partial derivatives. z=ln⁡(x2+y2)"

cistG

Answered question

2021-04-10

Find both first partial derivatives.

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

Answer & Explanation

jlo2niT

Skilled2021-04-12Added 96 answers

Step 1
Given function is

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

.
Partial derivative means differentiating with respect to one variable keeping other variable as constant.
Finding partial derivative of given function with respect to x, keeping y as constant.

z_{x} = \frac{\partial}{\partial x} [\ln (x^{2} + y^{2})]

= \frac{1}{(x^{2} + y^{2})} \cdot \frac{\partial}{\partial z} (x^{2} + y^{2})

= \frac{1}{(x^{2} + y^{2})} \cdot (2 x + 0)

= \frac{2 x}{x^{2} + y^{2}}

Step 2
Now, finding the partial derivative of given function with respect to y, keeping x as constant.

z_{y} = \frac{\partial}{\partial z} [\ln (x^{2} + y^{2})]

= \frac{1}{(x^{2} + y^{2})} \cdot \frac{\partial}{\partial y} (x^{2} + y^{2})

= \frac{1}{(x^{2} + y^{2})} \cdot (0 + 2 y)

= \frac{2 y}{x^{2} + y^{2}}

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Find both first partial derivatives. z = \ln(x^{2}+y^{2})

Answered question

Answer & Explanation

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