If `y = "log "((1 + sqrtx)/(1 - sqrtx)) " then " (dy)/(dx) =` ?

Options

A. `(1)/(sqrtx (1-x))`
B. `(-1)/(x(1 - sqrtx)^(2))`
C. `(-sqrtx)/(2(1 - sqrtx))`
D. none of these

Explanation

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

### Step 1: Apply the Logarithmic Property
Using the property of logarithms, we can rewrite the function:
\[ y = \log(1 + \sqrt{x}) - \log(1 - \sqrt{x}) \]

### Step 2: Differentiate Each Term
Now, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}[\log(1 + \sqrt{x})] - \frac{d}{dx}[\log(1 - \sqrt{x})] \]

### Step 3: Use the Chain Rule
Using the chain rule for differentiation, we find the derivatives:
\[ \frac{d}{dx}[\log(1 + \sqrt{x})] = \frac{1}{1 + \sqrt{x}} \cdot \frac{d}{dx}(1 + \sqrt{x}) = \frac{1}{1 + \sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \]
\[ \frac{d}{dx}[\log(1 - \sqrt{x})] = \frac{1}{1 - \sqrt{x}} \cdot \frac{d}{dx}(1 - \sqrt{x}) = \frac{1}{1 - \sqrt{x}} \cdot \left(-\frac{1}{2\sqrt{x}}\right) \]

### Step 4: Substitute Back into the Derivative
Substituting these back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{(1 + \sqrt{x}) \cdot 2\sqrt{x}} - \left(-\frac{1}{(1 - \sqrt{x}) \cdot 2\sqrt{x}}\right) \]
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}(1 + \sqrt{x})} + \frac{1}{2\sqrt{x}(1 - \sqrt{x})} \]

### Step 5: Combine the Fractions
Now, we combine the fractions:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \left(\frac{1}{1 + \sqrt{x}} + \frac{1}{1 - \sqrt{x}}\right) \]
Finding a common denominator for the fractions inside the parentheses:
\[ \frac{1}{1 + \sqrt{x}} + \frac{1}{1 - \sqrt{x}} = \frac{(1 - \sqrt{x}) + (1 + \sqrt{x})}{(1 + \sqrt{x})(1 - \sqrt{x})} = \frac{2}{1 - x} \]

### Step 6: Final Simplification
Substituting this back into our expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \cdot \frac{2}{1 - x} = \frac{1}{\sqrt{x}(1 - x)} \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x}(1 - x)} \]

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