If `y=tan^(-1)((sqrtx-x)/(1+x^(3//2)))`, then y'(1) is equal to:

Options

A. 0
B. `1/2`
C. `-1`
D. `-1/4`

Answer

To find \( y'(1) \) for the function \( y = \tan^{-1}\left(\frac{\sqrt{x} - x}{1 + x^{3/2}}\right) \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ y = \tan^{-1}\left(\frac{\sqrt{x} - x}{1 + x^{3/2}}\right) \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \), we use the formula for the derivative of the inverse tangent function:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{\sqrt{x} - x}{1 + x^{3/2}} \).

### Step 3: Differentiate \( u \)
Now, we need to differentiate \( u \):
\[ u = \frac{\sqrt{x} - x}{1 + x^{3/2}} \]
We will use the quotient rule:
\[ \frac{du}{dx} = \frac{(1 + x^{3/2})\frac{d}{dx}(\sqrt{x} - x) - (\sqrt{x} - x)\frac{d}{dx}(1 + x^{3/2})}{(1 + x^{3/2})^2} \]

### Step 4: Calculate \( \frac{d}{dx}(\sqrt{x} - x) \)
\[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}, \quad \frac{d}{dx}(-x) = -1 \]
Thus,
\[ \frac{d}{dx}(\sqrt{x} - x) = \frac{1}{2\sqrt{x}} - 1 \]

### Step 5: Calculate \( \frac{d}{dx}(1 + x^{3/2}) \)
\[ \frac{d}{dx}(1 + x^{3/2}) = \frac{3}{2}x^{1/2} \]

### Step 6: Substitute back into \( \frac{du}{dx} \)
Now substituting back:
\[ \frac{du}{dx} = \frac{(1 + x^{3/2})\left(\frac{1}{2\sqrt{x}} - 1\right) - (\sqrt{x} - x)\left(\frac{3}{2}x^{1/2}\right)}{(1 + x^{3/2})^2} \]

### Step 7: Evaluate \( \frac{dy}{dx} \) at \( x = 1 \)
Now we need to evaluate \( \frac{dy}{dx} \) at \( x = 1 \):
1. Calculate \( u \) at \( x = 1 \):
\[ u = \frac{\sqrt{1} - 1}{1 + 1^{3/2}} = \frac{1 - 1}{1 + 1} = 0 \]
2. Calculate \( 1 + u^2 \):
\[ 1 + u^2 = 1 + 0^2 = 1 \]
3. Now substitute \( x = 1 \) into \( \frac{du}{dx} \):
\[ \frac{du}{dx} \text{ at } x = 1 = \frac{(1 + 1)\left(\frac{1}{2} - 1\right) - (1 - 1)\left(\frac{3}{2}\right)}{(1 + 1)^2} \]
Simplifying this gives:
\[ \frac{du}{dx} = \frac{2 \cdot (-\frac{1}{2})}{4} = -\frac{1}{4} \]

### Step 8: Final calculation of \( y' \)
Now substituting back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{1} \cdot \left(-\frac{1}{4}\right) = -\frac{1}{4} \]

### Conclusion
Thus, we find that:
\[ y'(1) = -\frac{1}{4} \]