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

Short Answer

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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 \( f_y = \frac{-4y(y^2)}{(x^2 + y^2)^2} \)
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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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Most popular questions from this chapter

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