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Problem 10

# Find the first partial derivatives of the function. $$f(x, y)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$$

Expert verified
The first partial derivatives of the function $$f(x, y) = \frac{x^2 - y^2}{x^2 + y^2}$$ are: $$f_x = \frac{4x(y^2)}{(x^2 + y^2)^2}$$ and $$f_y = \frac{-4y(y^2)}{(x^2 + y^2)^2}$$
See the step by step solution

## Step 1: Find partial derivative with respect to x ($$f_x$$)

To find the partial derivative of $$f(x, y)$$ with respect to x, we'll apply the quotient rule for differentiation: $$f_x = \frac{(2x)(x^2 + y^2) - (x^2 - y^2)(2x)}{(x^2 + y^2)^2}$$

## Step 2: Simplify $$f_x$$

Now, let's simplify the expression for $$f_x$$: $$f_x = \frac{2x(x^2 + y^2) - 2x(x^2 - y^2)}{(x^2 + y^2)^2} \\ f_x = \frac{2x(x^2 + y^2 - x^2 + y^2)}{(x^2 + y^2)^2} \\ f_x = \frac{4x(y^2)}{(x^2 + y^2)^2}$$

## Step 3: Find partial derivative with respect to y ($$f_y$$)

Now let's find the partial derivative of $$f(x, y)$$ with respect to y, using the quotient rule again: $$f_y = \frac{(-2y)(x^2 + y^2) - (x^2 - y^2)(2y)}{(x^2 + y^2)^2}$$

## Step 4: Simplify $$f_y$$

Finally, let's simplify the expression for $$f_y$$: $$f_y = \frac{-2y(x^2 + y^2) - 2y(x^2 - y^2)}{(x^2 + y^2)^2} \\ f_y = \frac{-2y(x^2 + y^2 - x^2 + y^2)}{(x^2 + y^2)^2} \\ f_y = \frac{-4y(y^2)}{(x^2 + y^2)^2}$$ So, the first partial derivatives of the given function are: $$f_x = \frac{4x(y^2)}{(x^2 + y^2)^2}$$ and $$f_y = \frac{-4y(y^2)}{(x^2 + y^2)^2}$$

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