Find the derivative of the following w.r.t. x :
`tan(sin^(-1)x)`

Answer

To find the derivative of \( y = \tan(\sin^{-1} x) \) with respect to \( x \), we will apply the chain rule. Here are the steps to solve the problem:

### Step 1: Define the function
Let \( y = \tan(\sin^{-1} x) \).

### Step 2: Differentiate using the chain rule
To differentiate \( y \) with respect to \( x \), we will use the chain rule. The derivative of \( \tan(u) \) with respect to \( u \) is \( \sec^2(u) \), where \( u = \sin^{-1} x \).

Thus, we have:
\[ \frac{dy}{dx} = \sec^2(\sin^{-1} x) \cdot \frac{d}{dx}(\sin^{-1} x) \]

### Step 3: Find the derivative of \( \sin^{-1} x \)
The derivative of \( \sin^{-1} x \) is given by:
\[ \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \]

### Step 4: Substitute back into the derivative
Now substituting this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \sec^2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1 - x^2}} \]

### Step 5: Simplify \( \sec^2(\sin^{-1} x) \)
To simplify \( \sec^2(\sin^{-1} x) \), we can use the identity:
\[ \sec^2(\theta) = 1 + \tan^2(\theta) \]
where \( \theta = \sin^{-1} x \).

From the definition of \( \sin^{-1} x \), we have:
- \( \sin(\theta) = x \)
- Using the Pythagorean identity, \( \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2} \)

Thus,
\[ \tan(\sin^{-1} x) = \frac{\sin(\sin^{-1} x)}{\cos(\sin^{-1} x)} = \frac{x}{\sqrt{1 - x^2}} \]

Then,
\[ \tan^2(\sin^{-1} x) = \frac{x^2}{1 - x^2} \]

So,
\[ \sec^2(\sin^{-1} x) = 1 + \tan^2(\sin^{-1} x) = 1 + \frac{x^2}{1 - x^2} = \frac{1 - x^2 + x^2}{1 - x^2} = \frac{1}{1 - x^2} \]

### Step 6: Substitute \( \sec^2(\sin^{-1} x) \) back
Now substituting this back into our derivative:
\[ \frac{dy}{dx} = \frac{1}{1 - x^2} \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{1}{(1 - x^2)^{3/2}} \]

### Final Answer
Thus, the derivative of \( y = \tan(\sin^{-1} x) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{(1 - x^2)^{3/2}} \]