Question

Find the value of \(\frac{dy}{dx}\) if \(y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)\), where \(-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\).

Hint:

Whenever you see expressions like \(\frac{2x}{1-x^2}\), \(\frac{1-x^2}{1+x^2}\), \(\frac{3x-x^3}{1-3x^2}\) inside inverse trigonometric functions, always consider substituting \(x=\tan\theta\). Similarly, for expressions involving \(\sqrt{1-x^2}\), try \(x=\sin\theta\) or \(x=\cos\theta\). This often simplifies the problem dramatically.

Answer 1

Step 1: Understanding the Concept:
The problem asks for the derivative of an inverse trigonometric function. The expression inside the \(\tan^{-1}\) function is a hint to use a trigonometric substitution to simplify the function before differentiating. The expression is identical in form to the triple angle formula for tangent.
Step 2: Key Formula or Approach:
1. Recognize the trigonometric identity: \(\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\).
2. Use the substitution \(x = \tan\theta\), which implies \(\theta = \tan^{-1}x\).
3. Use the given domain of \(x\) to find the range of \(\theta\).
4. Simplify the expression for \(y\) using the identity and the property \(\tan^{-1}(\tan z) = z\) (within the principal value range).
5. Differentiate the simplified expression for \(y\) with respect to \(x\).
Step 3: Detailed Explanation or Calculation:
Let \(x = \tan\theta\). Then \(\theta = \tan^{-1}x\).
The given domain for \(x\) is \(-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\).
Substituting \(x = \tan\theta\), we get \(-\frac{1}{\sqrt{3}}<\tan\theta<\frac{1}{\sqrt{3}}\).
This implies \(-\frac{\pi}{6}<\theta<\frac{\pi}{6}\).
Now, substitute \(x = \tan\theta\) into the expression for \(y\): \[ y = \tan^{-1}\left(\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\right) \] Using the triple angle identity, this simplifies to: \[ y = \tan^{-1}(\tan(3\theta)) \] To simplify this further, we must check if \(3\theta\) lies within the principal value range of \(\tan^{-1}\), which is \((-\frac{\pi}{2}, \frac{\pi}{2})\).
From our range for \(\theta\), \(-\frac{\pi}{6}<\theta<\frac{\pi}{6}\), we can find the range for \(3\theta\): \[ 3 \times (-\frac{\pi}{6})<3\theta<3 \times (\frac{\pi}{6}) \implies -\frac{\pi}{2}<3\theta<\frac{\pi}{2} \] Since \(3\theta\) is within the principal value range, we can write: \[ y = 3\theta \] Now, substitute back \(\theta = \tan^{-1}x\): \[ y = 3\tan^{-1}x \] Finally, differentiate \(y\) with respect to \(x\): \[ \frac{dy}{dx} = \frac{d}{dx}(3\tan^{-1}x) = 3 \cdot \frac{1}{1+x^2} = \frac{3}{1+x^2} \] Step 4: Final Answer:
The value of \(\frac{dy}{dx}\) is \(\frac{3}{1+x^2}\).