2.5 Q-12

Question Statement

We are given the function: $y = \tan \left( \tan^{-1} x \right)$

We are tasked with proving the following identity: $\left( 1 + x^2 \right) y_1 - p \left( 1 + y^2 \right) = 0$

where $y_1$ denotes the derivative of $y$ with respect to $x$.

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Background and Explanation

To solve this problem, we will need to apply basic differentiation rules, including the chain rule, to find the derivative of $y$ with respect to $x$. The function involves the inverse tangent function ($\tan^{-1}(x)$), and we need to differentiate it within the given tangent expression. Understanding how to differentiate composite functions is essential for solving this problem.

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Solution

Step 1: Express $y$

The given equation is: $y = \tan \left( \tan^{-1} x \right)$

This simplifies to $y = x$, because: $\tan \left( \tan^{-1} x \right) = x$

Thus, we now know: $y = x$

Step 2: Differentiate $y$

Now, we differentiate both sides of the equation with respect to $x$: $\frac{d y}{d x} = \frac{d}{d x} \left( \tan \left( \tan^{-1} x \right) \right)$

Using the chain rule, we differentiate the right-hand side: $\frac{d y}{d x} = \sec^2 \left( \tan^{-1} x \right) \cdot \frac{d}{d x} \left( \tan^{-1} x \right)$

Now, recall the derivative of $\tan^{-1}(x)$: $\frac{d}{d x} \left( \tan^{-1} x \right) = \frac{1}{1 + x^2}$

Thus, we get: $\frac{d y}{d x} = \sec^2 \left( \tan^{-1} x \right) \cdot \frac{1}{1 + x^2}$

Step 3: Simplify $\sec^2 \left( \tan^{-1} x \right)$

From trigonometric identities, we know: $\sec^2 \theta = 1 + \tan^2 \theta$

So, $\sec^2 \left( \tan^{-1} x \right) = 1 + x^2$

Substituting this into the derivative, we have: $\frac{d y}{d x} = (1 + x^2) \cdot \frac{1}{1 + x^2}$

This simplifies to: $\frac{d y}{d x} = 1$

Step 4: Set up the equation to prove

We need to prove: $\left( 1 + x^2 \right) y_1 - p \left( 1 + y^2 \right) = 0$

From the previous steps, we know that $y_1 = 1$, and $y = x$, so: $\left( 1 + x^2 \right) \cdot 1 - p \left( 1 + x^2 \right) = 0$

This simplifies to: $\left( 1 + x^2 \right) - p \left( 1 + x^2 \right) = 0$

Factoring out $(1 + x^2)$, we get: $(1 + x^2) \left( 1 - p \right) = 0$

Step 5: Conclusion

For the equation to hold true, we must have: $1 - p = 0$

Therefore, $p = 1$, which completes the proof.

Hence, the identity is proven as: $\left( 1 + x^2 \right) y_1 - p \left( 1 + y^2 \right) = 0$

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Key Formulas or Methods Used

1. Derivative of Inverse Tangent: $\frac{d}{dx} \left( \tan^{-1} x \right) = \frac{1}{1 + x^2}$

2. Trigonometric Identity: $\sec^2 \theta = 1 + \tan^2 \theta$

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Summary of Steps

1. Simplify $y = \tan \left( \tan^{-1} x \right)$ to $y = x$.
2. Differentiate $y = x$ to get $y_1 = 1$.
3. Set up the equation $\left( 1 + x^2 \right) y_1 - p \left( 1 + y^2 \right) = 0$.
4. Simplify the equation to $(1 + x^2) (1 - p) = 0$.
5. Conclude that $p = 1$, proving the identity.

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