Differentiate `Sin^(-1) [xsqrtx]` w.r.t.x.

Answer

To differentiate the function \( \sin^{-1}(x \sqrt{x}) \) with respect to \( x \), we will follow these steps:

### Step 1: Identify the function
Let \( h = \sin^{-1}(x \sqrt{x}) \).

### Step 2: Differentiate using the chain rule
The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is given by:
\[ \frac{d}{du} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \]
Here, \( u = x \sqrt{x} \).

### Step 3: Find \( \frac{du}{dx} \)
Now we need to differentiate \( u = x \sqrt{x} \):
\[ u = x \cdot x^{1/2} = x^{3/2} \]
Using the power rule:
\[ \frac{du}{dx} = \frac{3}{2} x^{1/2} \]

### Step 4: Apply the chain rule
Using the chain rule, we have:
\[ \frac{dh}{dx} = \frac{dh}{du} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1 - (x \sqrt{x})^2}} \cdot \frac{du}{dx} \]

### Step 5: Substitute \( u \) and \( \frac{du}{dx} \)
Now substitute \( u \) and \( \frac{du}{dx} \):
\[ \frac{dh}{dx} = \frac{1}{\sqrt{1 - (x^{3/2})^2}} \cdot \frac{3}{2} x^{1/2} \]
This simplifies to:
\[ \frac{dh}{dx} = \frac{1}{\sqrt{1 - x^3}} \cdot \frac{3}{2} x^{1/2} \]

### Step 6: Final expression
Thus, the derivative of \( \sin^{-1}(x \sqrt{x}) \) with respect to \( x \) is:
\[ \frac{dh}{dx} = \frac{3 \sqrt{x}}{2 \sqrt{1 - x^3}} \]