Question

If \( y = \log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 \), then show that \( x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2 \).

Hint:

For logarithmic differentiation, simplify using logarithm properties before differentiating.

Answer 1

To solve the problem, we are given:
\[ y = \log \left( \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 \right) \] We are to prove: \[ x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2 \]

1. Simplify the Expression:
Use the identity \( \log(a^2) = 2\log a \):
\[ y = 2 \log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) \]

2. Let’s Define:
Let \( u = \sqrt{x} + \frac{1}{\sqrt{x}} \)
Then \( y = 2 \log u \)

3. Differentiate First Time (First Derivative \( y_1 \)):
Using the chain rule:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{u} \cdot \frac{du}{dx} \] Now compute \( \frac{du}{dx} \):
\[ \frac{d}{dx} \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} = \frac{1}{2\sqrt{x}} \left( 1 - \frac{1}{x} \right) \] So: \[ y_1 = \frac{2}{\sqrt{x} + \frac{1}{\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \left(1 - \frac{1}{x} \right) \]

Simplify numerator and denominator: Let’s instead substitute back and simplify using an alternate route.

Alternate Simpler Substitution:
\[ y = \log \left( \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 \right) = \log \left( x + \frac{1}{x} + 2 \right) = \log \left( \frac{(x + 1)^2}{x} \right) \] So: \[ y = \log \left( \frac{(x + 1)^2}{x} \right) = \log (x + 1)^2 - \log x = 2 \log(x + 1) - \log x \]

4. First Derivative \( y_1 = \frac{dy}{dx} \):
\[ y_1 = 2 \cdot \frac{1}{x + 1} - \frac{1}{x} \]

5. Second Derivative \( y_2 = \frac{d^2y}{dx^2} \):
\[ y_2 = \frac{d}{dx} \left( \frac{2}{x + 1} - \frac{1}{x} \right) = -\frac{2}{(x + 1)^2} + \frac{1}{x^2} \]

6. Substitute in the Given Expression:
We are to show: \[ x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2 \] Substitute \( y_1 \) and \( y_2 \):

\[ x(x + 1)^2 \left( -\frac{2}{(x + 1)^2} + \frac{1}{x^2} \right) + (x + 1)^2 \left( \frac{2}{x + 1} - \frac{1}{x} \right) \]

Simplify both terms:

First term:
\[ x(x + 1)^2 \left( -\frac{2}{(x + 1)^2} + \frac{1}{x^2} \right) = x \left( -2 + \frac{(x + 1)^2}{x^2} \right) \]

Second term:
\[ (x + 1)^2 \left( \frac{2}{x + 1} - \frac{1}{x} \right) = (x + 1) \cdot 2 - \frac{(x + 1)^2}{x} \]

Add both together: \[ x \left( -2 + \frac{(x + 1)^2}{x^2} \right) + \left[ 2(x + 1) - \frac{(x + 1)^2}{x} \right] \]

Combine the expressions: \[ -2x + \frac{(x + 1)^2}{x} + 2(x + 1) - \frac{(x + 1)^2}{x} = -2x + 2(x + 1) = -2x + 2x + 2 = 2 \]

Final Answer:
\[ x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2 \quad \text{is proved.} \]