Step by step solution to "Find all first partial derivatives. z=ln⁡(x2+y2+1)"

CheemnCatelvew

Answered question

2021-02-21

Find all first partial derivatives.

z = \ln (x^{2} + y^{2} + 1)

insonsipthinye

Skilled2021-02-22Added 83 answers

Step 1
Given
The equation is

z = \ln (x^{2} + y^{2} + 1)

.
Step 2
To find all first partial derivatives .
To find

\frac{d z}{d x}

\frac{d z}{d x} (\ln (x^{2} + y^{2} + 1))

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

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

\frac{d z}{d x} = \frac{2 x}{x^{2} + y^{2} + 1}

To find

\frac{d z}{d y}

\frac{d z}{d y} = \frac{d}{d y} (\ln (x^{2} + y^{2} + 1))

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

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

\frac{d z}{d y} = \frac{2 y}{x^{2} + y^{2} + 1}

Therefore , The all partial derivatives are

\frac{d z}{d x} = \frac{2 x}{x^{2} + y^{2} + 1} and \frac{d z}{d y} = \frac{2 y}{x^{2} + y^{2} + 1}

Jeffrey Jordon

Expert2021-11-17Added 2605 answers

Answer is given below (on video)

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