Differentiate: `tan^(-1)""(sqrt(1-x^2))/x` with respect to `cos^(-1)(2xsqrt(1-x^2))` , when `x ne0`

Answer

To differentiate \( y_1 = \tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \) with respect to \( y_2 = \cos^{-1}(2x\sqrt{1-x^2}) \), we will follow these steps:

### Step 1: Substitute \( x \) with \( \cos \theta \)
Let \( x = \cos \theta \). Then, we have:
\[ \sqrt{1 - x^2} = \sqrt{1 - \cos^2 \theta} = \sin \theta \]

### Step 2: Rewrite \( y_1 \) and \( y_2 \)
Now we can rewrite \( y_1 \) and \( y_2 \):
\[ y_1 = \tan^{-1}\left(\frac{\sin \theta}{\cos \theta}\right) = \tan^{-1}(\tan \theta) = \theta \quad \text{(for } -\frac{\pi}{2} < \theta < \frac{\pi}{2}\text{)} \]
\[ y_2 = \cos^{-1}(2\cos \theta \sin \theta) = \cos^{-1}(\sin 2\theta) \]

### Step 3: Differentiate \( y_1 \) with respect to \( \theta \)
The derivative of \( y_1 \) with respect to \( \theta \) is:
\[ \frac{dy_1}{d\theta} = 1 \]

### Step 4: Differentiate \( y_2 \) with respect to \( \theta \)
Using the chain rule:
\[ \frac{dy_2}{d\theta} = \frac{d}{d\theta} \cos^{-1}(\sin 2\theta) = -\frac{1}{\sqrt{1 - (\sin 2\theta)^2}} \cdot \frac{d}{d\theta}(\sin 2\theta) \]
\[ = -\frac{1}{\sqrt{1 - \sin^2 2\theta}} \cdot 2\cos 2\theta = -\frac{2\cos 2\theta}{\sqrt{\cos^2 2\theta}} = -2\cos 2\theta \]

### Step 5: Find \( \frac{dy_1}{dy_2} \)
Now we can find \( \frac{dy_1}{dy_2} \):
\[ \frac{dy_1}{dy_2} = \frac{\frac{dy_1}{d\theta}}{\frac{dy_2}{d\theta}} = \frac{1}{-2\cos 2\theta} = -\frac{1}{2\cos 2\theta} \]

### Final Answer
Thus, the derivative of \( y_1 \) with respect to \( y_2 \) is:
\[ \frac{dy_1}{dy_2} = -\frac{1}{2\cos 2\theta} \]