Question

If \(f(x, y)=\tan ^{-1}\left(y^{2} / x\right)\), find \(f_{x}(\sqrt{5},-2)\) and \(f_{y}(\sqrt{5},-2)\)

Short answer

\( f_x(\sqrt{5}, -2) = -\frac{4}{21} \) and \( f_y(\sqrt{5}, -2) = -\frac{4}{21} \).

Step-by-step solution

Understanding the Partial Derivatives

To find the partial derivatives of the function \( f(x, y) = an^{-1}(y^2 / x) \), we need to understand that \( f_x \) and \( f_y \) represent the derivatives of \( f \) with respect to \( x \) and \( y \), respectively. This involves differentiating with respect to one variable while treating the other as constant.

Finding the Partial Derivative with respect to x, \( f_x \)

The function is \( f(x, y) = an^{-1}rac{y^2}{x} \). Using the chain rule, the derivative with respect to \( x \) is \( f_x = \frac{d}{dx}\left(\tan^{-1}\left(\frac{y^2}{x}\right)\right) = \frac{1}{1 + (y^2/x)^2} \cdot \left(-\frac{y^2}{x^2}\right) \). Thus, \( f_x = -\frac{y^2}{x^2 + y^4} \).

Calculating \( f_x(\sqrt{5}, -2) \)

Substitute \( x = \sqrt{5} \) and \( y = -2 \) into the expression for \( f_x \). Thus, \( f_x(\sqrt{5}, -2) = -\frac{(-2)^2}{(\sqrt{5})^2 + (-2)^4} = -\frac{4}{5 + 16} = -\frac{4}{21} \).

Finding the Partial Derivative with respect to y, \( f_y \)

Using the chain rule again, \( f_y = \frac{d}{dy}\left(\tan^{-1}\left(\frac{y^2}{x}\right)\right) = \frac{1}{1 + (y^2/x)^2} \cdot \left(\frac{2y}{x}\right) \). Therefore, \( f_y = \frac{2y}{x^2 + y^4} \).

Calculating \( f_y(\sqrt{5}, -2) \)

Substitute \( x = \sqrt{5} \) and \( y = -2 \) into the expression for \( f_y \). Thus, \( f_y(\sqrt{5}, -2) = \frac{2(-2)}{(\sqrt{5})^2 + (-2)^4} = \frac{-4}{5 + 16} = \frac{-4}{21} \).

Key concepts

Chain Rule

The chain rule is a crucial tool in calculus, particularly when dealing with composite functions. It allows us to find the derivative of a function that is nested within another function. For instance, if we have a function like \( f(x, y) = \tan^{-1}(y^2 / x) \), it involves finding the derivative of the inverse tangent function with a compound argument \( y^2 / x \).

When applying the chain rule, we differentiate the outer function first, keeping the inner function unchanged. Next, we multiply by the derivative of the inner function. This two-step process helps us break down complex derivatives into simpler parts.

• Differentiate the outer layer (\( \tan^{-1} \)) to get \( \frac{1}{1 + (u)^2} \), with \( u = \frac{y^2}{x} \).
• Differentiate the inner function, \( u = y^2 / x \), to get \( -\frac{y^2}{x^2} \) when considering \( u \) with respect to \( x \), and \( \frac{2y}{x} \) with respect to \( y \).

This rule is vital for untangling nested functions and is frequently used in partial derivatives, as it was applied in finding \( f_x \) and \( f_y \).

Derivative with respect to x

The partial derivative with respect to \( x \) of a function \( f(x, y) \) involves treating \( y \) as a constant and differentiating the entire function as if it were a single-variable function in terms of \( x \).

For our function \( f(x, y) = \tan^{-1}(y^2 / x) \), using the chain rule, we compute the derivative with respect to \( x \), resulting in:

• First, find the outer function's derivative: \( \frac{1}{1 + (y^2/x)^2} \).
• Next, multiply by the derivative of the inner function with respect to \( x \): \( -\frac{y^2}{x^2} \).
• This gives a final result of: \[ f_x = -\frac{y^2}{x^2 + y^4} \].

When evaluating at a specific point, such as \( x = \sqrt{5} \) and \( y = -2 \), substitute directly into the expression to get the final result \( -\frac{4}{21} \).

Derivative with respect to y

Like the derivative with respect to \( x \), the partial derivative with respect to \( y \) involves treating \( x \) as a constant. This allows us to simplify the differentiation of a function \( f(x, y) \) by only considering changes along the \( y \)-axis, keeping \( x \) fixed.

For the function \( f(x, y) = \tan^{-1}(y^2 / x) \), the steps include:

• Use the chain rule to differentiate the outer function: \( \frac{1}{1 + (y^2/x)^2} \).
• Multiply by the derivative of the inner function \( y^2 / x \) with respect to \( y \): \( \frac{2y}{x} \).
• Thus, the partial derivative with respect to \( y \) becomes \[ f_y = \frac{2y}{x^2 + y^4} \].

Substituting the specific point \( x = \sqrt{5} \) and \( y = -2 \) gives us the intuitive result of \( \frac{-4}{21} \).

Tangent Inverse Function

The inverse tangent function, written as \( \tan^{-1}(x) \), is an essential trigonometric function often encountered in calculus problems, especially involving derivatives and integrals.

Here, it appears in the form \( \tan^{-1}\left(\frac{y^2}{x}\right) \), where the argument is a fraction of two variables. The inverse tangent function's derivative is foundational in this context:

• It is given by the formula \( \frac{d}{dx}\left(\tan^{-1}(u)\right) = \frac{1}{1+u^2}\).
• This general formula is applied regardless of what \( u \) is, helping us decompose complex functions like \( \frac{y^2}{x} \) effectively.

In problems requiring partial derivatives, recognizing how the inverse tangent function behaves is key, making it easier to apply the chain rule and find the necessary solutions effectively.