If `y = (u -1)/(u + 1) and u = sqrt(x)`, then `(dy)/(dx)` is

Options

A. `1/(sqrt(x)(1 + sqrt(x))^(2))`
B. `1/(sqrt(x)(1-sqrt(x))^(2))`
C. `1/(sqrt(x)(1 + sqrt(x))^(-2))`
D. `sqrt(x)(1 + sqrt(x))^(2)`

Answer

To find \(\frac{dy}{dx}\) given \(y = \frac{u - 1}{u + 1}\) and \(u = \sqrt{x}\), we will use the chain rule for differentiation. Here are the steps to solve the problem:

### Step 1: Substitute \(u\) into the equation for \(y\)
We start by substituting \(u = \sqrt{x}\) into the equation for \(y\):
\[ y = \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \]

### Step 2: Differentiate \(y\) with respect to \(u\)
Next, we differentiate \(y\) with respect to \(u\) using the quotient rule. The quotient rule states that if \(y = \frac{f(u)}{g(u)}\), then:
\[ \frac{dy}{du} = \frac{f'(u)g(u) - f(u)g'(u)}{(g(u))^2} \]
Here, \(f(u) = u - 1\) and \(g(u) = u + 1\).

Calculating the derivatives:
- \(f'(u) = 1\)
- \(g'(u) = 1\)

Now applying the quotient rule:
\[ \frac{dy}{du} = \frac{(1)(u + 1) - (u - 1)(1)}{(u + 1)^2} \]
This simplifies to:
\[ \frac{dy}{du} = \frac{(u + 1) - (u - 1)}{(u + 1)^2} = \frac{u + 1 - u + 1}{(u + 1)^2} = \frac{2}{(u + 1)^2} \]

### Step 3: Differentiate \(u\) with respect to \(x\)
Now we differentiate \(u = \sqrt{x}\) with respect to \(x\):
\[ \frac{du}{dx} = \frac{1}{2\sqrt{x}} \]

### Step 4: Apply the chain rule to find \(\frac{dy}{dx}\)
Now we can apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
Substituting the values we found:
\[ \frac{dy}{dx} = \frac{2}{(u + 1)^2} \cdot \frac{1}{2\sqrt{x}} \]
Substituting \(u = \sqrt{x}\):
\[ \frac{dy}{dx} = \frac{2}{(\sqrt{x} + 1)^2} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}(\sqrt{x} + 1)^2} \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x}(\sqrt{x} + 1)^2} \]