Solved: "Find the first partial derivatives of the following..."

Jaya Legge

Answered question

2021-02-21

Find the first partial derivatives of the following functions.

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

SabadisO

Skilled2021-02-23Added 108 answers

Given,

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

Differentiating partially with respect to x, we get

f_{x} (x, y) = \frac{\frac{\partial (x^{2})}{\partial x} \cdot (x^{2} + y^{2}) - \frac{\partial (x^{2} + y^{2})}{\partial x} \cdot x^{2}}{{(x^{2} + y^{2})}^{2}} [∵ \frac{d}{d x} (\frac{u}{v}) = \frac{\frac{d u}{d x} \cdot v - \frac{d v}{d x} \cdot u}{v^{2}}]

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

= \frac{2 x^{3} + 2 x y^{2} - 2 x^{3}}{{(x^{2} + y^{2})}^{2}}

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

Step 2
Differentiating partially with respect to y, we get

f_{y} (x, y) = \frac{\frac{\partial (x^{2})}{\partial y} \cdot (x^{2} + y^{2}) - \frac{\partial (x^{2} + y^{2})}{\partial y} \cdot x^{2}}{{(x^{2} + y^{2})}^{2}} [∵ \frac{d}{d x} (\frac{u}{v}) = \frac{\frac{d u}{d x} \cdot v - \frac{d v}{d x} \cdot u}{v^{2}}]

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

This is helpful 6 Flag

Jeffrey Jordon

Expert2022-01-31Added 2605 answers

Answer is given below (on video)

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