2.5 Q-11

Question Statement

Given that: $\frac{y}{x} = \tan^{-1} \left( \frac{y}{x} \right)$

We are tasked with proving that: $\frac{d y}{d x} = \frac{y}{x}$

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Background and Explanation

This problem involves differentiation using implicit differentiation and applying the chain rule. It assumes familiarity with inverse trigonometric functions, particularly $\tan^{-1}(u)$, and basic differentiation rules. We also need to handle derivatives involving both $x$ and $y$ as functions of $x$, which is where implicit differentiation comes into play.

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Solution

To solve this, we will differentiate both sides of the given equation with respect to $x$ and simplify the terms. Let’s break it down step-by-step:

1. Start with the given equation: $\frac{y}{x} = \tan^{-1} \left( \frac{y}{x} \right)$

2. Differentiate both sides with respect to $x$:

   Using implicit differentiation on both sides: $\frac{d}{dx} \left( \frac{y}{x} \right) = \frac{d}{dx} \left( \tan^{-1} \left( \frac{y}{x} \right) \right)$

3. Apply the quotient rule and chain rule:

   On the left-hand side, use the quotient rule: $\frac{d}{dx} \left( \frac{y}{x} \right) = \frac{x \frac{d y}{d x} - y}{x^2}$

   On the right-hand side, use the chain rule for differentiating $\tan^{-1}(u)$, where $u = \frac{y}{x}$: $\frac{d}{dx} \left( \tan^{-1} \left( \frac{y}{x} \right) \right) = \frac{1}{1 + \left( \frac{y}{x} \right)^2} \cdot \frac{d}{dx} \left( \frac{y}{x} \right)$

   We need to differentiate $\frac{y}{x}$, which gives us: $\frac{d}{dx} \left( \frac{y}{x} \right) = \frac{x \frac{d y}{d x} - y}{x^2}$

4. Substitute into the equation:

   Now, substitute these expressions back into the equation: $\frac{x \frac{d y}{d x} - y}{x^2} = \frac{1}{1 + \left( \frac{y}{x} \right)^2} \cdot \frac{x \frac{d y}{d x} - y}{x^2}$

5. Simplify the terms:

   Notice that both sides of the equation have the term $\frac{x \frac{d y}{d x} - y}{x^2}$. Simplifying this gives: $1 = \frac{1}{1 + \left( \frac{y}{x} \right)^2}$

   Now, let’s work with the denominator: $1 + \left( \frac{y}{x} \right)^2 = \frac{x^2 + y^2}{x^2}$

   Therefore, the equation becomes: $\frac{x^2 + y^2}{x^2} = 1$

   This simplifies to: $x^2 + y^2 = x^2$

6. Final Simplification:

   We now see that the left-hand side and right-hand side are equal, which confirms that: $\frac{d y}{d x} = \frac{y}{x}$

Thus, we have proven that: $\frac{d y}{d x} = \frac{y}{x}$

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Key Formulas or Methods Used

1. Quotient Rule for differentiation: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

2. Chain Rule for differentiating composite functions: $\frac{d}{dx} \left( f(g(x)) \right) = f'(g(x)) \cdot g'(x)$

3. Derivative of the Inverse Tangent Function: $\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}$

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Summary of Steps

1. Start with the given equation: $\frac{y}{x} = \tan^{-1} \left( \frac{y}{x} \right)$.
2. Differentiate both sides with respect to $x$.
3. Apply the quotient rule on the left side and the chain rule on the right side.
4. Simplify the resulting expressions.
5. Show that the left-hand side and right-hand side are equal.
6. Conclude that $\frac{d y}{d x} = \frac{y}{x}$.

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