If ` y = tan ^-1[x/(1+sqrt(1-x^2))] "then find" dy/dx`

Explanation

To find \( \frac{dy}{dx} \) for the function \( y = \tan^{-1} \left( \frac{x}{1 + \sqrt{1 - x^2}} \right) \), we can follow these steps:

### Step 1: Simplify the Argument of the Inverse Tangent
Let \( x = \sin \theta \). Then, the expression inside the inverse tangent becomes:
\[ y = \tan^{-1} \left( \frac{\sin \theta}{1 + \sqrt{1 - \sin^2 \theta}} \right) \]

Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we have:
\[ \sqrt{1 - \sin^2 \theta} = \cos \theta \]
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