Problem 43 - Single Variable Calculus

Question

Use implicit differentiation to find \(d y / d x\). $$ \tan ^{-1}\left(\frac{y}{x}\right)-\ln \sqrt{x^{2}+y^{2}}=0 $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{y}{x^2} + \frac{(x^2+y^2) \tan\left(\frac{y}{x}\right)}{x^2} \)

Step 1: Differentiate both sides of the equation with respect to x

Using implicit differentiation, we will differentiate the given equation: \( \frac{d}{dx}[\tan^{-1}\left(\frac{y}{x}\right)] - \frac{d}{dx}[\ln\sqrt{x^2+y^2}] = \frac{d}{dx}[0] \)

Step 2: Apply the Chain Rule to differentiate the arctangent term

The Chain Rule states that \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \). We'll apply this rule to the first term, using the derivative of the arctangent and the quotient: \( \frac{1}{1+(\frac{y}{x})^2} \cdot \frac{d}{dx}\left(\frac{y}{x}\right) - \frac{d}{dx}[\ln\sqrt{x^2+y^2}] = 0 \)

Step 3: Differentiate the quotient term

Using the quotient rule, \( \frac{d}{dx}[\frac{u}{v}] = \frac{vu'-uv'}{v^2} \), differentiate the quotient term: \( \frac{1}{1+(\frac{y}{x})^2} \cdot \frac{x\frac{dy}{dx}-y}{x^2} - \frac{d}{dx}[\ln\sqrt{x^2+y^2}] = 0 \)

Step 4: Apply Chain Rule to differentiate the natural logarithm term

Now apply the Chain Rule to the second term: \( \frac{1}{1+(\frac{y}{x})^2} \cdot \frac{x\frac{dy}{dx}-y}{x^2} - \frac{1}{\sqrt{x^2+y^2}}\cdot\frac{d}{dx} [x^2+y^2] = 0 \)

Step 5: Differentiate the expression inside the square root

Differentiate the expression inside the square root, then simplify the equation: \( \frac{1}{1+(\frac{y}{x})^2} \cdot \frac{x\frac{dy}{dx}-y}{x^2} - \frac{1}{\sqrt{x^2+y^2}}\cdot(2x+0) = 0 \)

Step 6: Solve for dy/dx

Now, isolate dy/dx and solve for it: \( \frac{1}{1+(\frac{y}{x})^2} \cdot \frac{x\frac{dy}{dx}-y}{x^2} = \frac{1}{\sqrt{x^2+y^2}}\cdot(2x) \) \( \frac{dy}{dx} = \frac{y}{x^2} + \frac{(x^2+y^2) \tan\left(\frac{y}{x}\right)}{x^2}\) And this is the derivative of the given equation with respect to x.

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used to find the derivative of a composite function. It is essential when dealing with implicit differentiation, as seen in this exercise. A composite function is a function that is made up of two or more functions. For instance, if you have a function inside of another, like the arctangent function in our equation, the Chain Rule is the tool you use to differentiate it.

Here's a quick breakdown:

• You first differentiate the outer function while leaving the inner function unchanged.
• Then, multiply that result by the derivative of the inner function.

In the exercise, \( an^{-1}\left(\frac{y}{x}\right)\) is a composite function. Applying the Chain Rule, we differentiate \( an^{-1}(z)\) with respect to \(z\), resulting in \(\frac{1}{1+z^2}\). Then, multiply by the derivative of \(\frac{y}{x}\) using the Quotient Rule.

Quotient Rule

The Quotient Rule is specifically used for finding the derivatives of divisions of two functions. This rule states: if \(u(x)\) and \(v(x)\) are functions of \(x\), then the derivative of their quotient \(\frac{u}{v}\) is given by:

• \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2} \]

In our context, \(rac{y}{x}\) represents \(u/v\). When reliant on implicit differentiation where \(y\) itself is a function of \(x\), this means calculating the derivative involves using both the Chain Rule and the Quotient Rule. Here, the derivative is a step towards solving for \(\frac{dy}{dx}\).

Utilizing the Quotient Rule tied with implicit differentiation reveals a refined, systematic approach—first differentiating \(y\) as \(\frac{dy}{dx}\), acknowledging every factor within the division.

Inverse Trigonometric Functions

Inverse trigonometric functions like \( an^{-1}(x)\), also known as the arctangent function, are crucial in calculus for identifying angles based on their trigonometric ratios. When differentiating these functions implicitly, understanding their rates of change relative to their inputs is pivotal. The derivative of \(\tan^{-1}(x)\) with respect to \(x\) is \(\frac{1}{1+x^2}\).

In this problem, we apply this knowledge to \( an^{-1}\left(\frac{y}{x}\right)\). The derivative isn't straightforward since \(y/x\) is itself a function. Therefore, the Chain Rule and Quotient Rule adapt our differentiation process to evaluate these interconnected variables accurately.

Inverse trigonometric derivatives find numerous applications in dealing with angular transformations in real-world scenarios, enabling us to interpret complex rotations and angles.

Natural Logarithm Differentiation

Differentiating natural logarithms is an essential skill in calculus. The function \( ext{ln}(x)\) has the derivative \(rac{1}{x}\), which is particularly useful when combined with the Chain Rule for more complex expressions like \( ext{ln}\sqrt{x^2+y^2}\) in this exercise.

By applying the Chain Rule, we look inside the natural logarithm function (which involves a square root, \(\sqrt{x^2+y^2}\)). Start by acknowledging the derivative relation of \( ext{ln}(z)\) leading to \(\frac{1}{z}\), then shift focus to the inner portion, differentiating \(x^2+y^2\). This process is essential for properly calculating any adjustments due to the implicitly dependent variable \(y\).

Natural logarithm differentiation extends its use in various fields, from solving exponential growth problems to understanding decay rates, providing a bridge between algebraic expressions and real-world phenomena.