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

Given the equation:\[y = \log \left( \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 \right)\]We need to prove:\[x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2\]

1. Simplify the given expression:
Using the logarithmic property \( \log(a^2) = 2\log a \):\[y = 2 \log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)\]

2. Define a substitution:
Let \( u = \sqrt{x} + \frac{1}{\sqrt{x}} \). Then the expression becomes \( y = 2 \log u \).

3. Calculate the first derivative \( y_1 \):
Applying the chain rule:\[\frac{dy}{dx} = 2 \cdot \frac{1}{u} \cdot \frac{du}{dx}\]Now, we find \( \frac{du}{dx} \):\[\frac{du}{dx} = \frac{d}{dx} \left( x^{1/2} + x^{-1/2} \right) = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} = \frac{1}{2\sqrt{x}} \left( 1 - \frac{1}{x} \right)\]Substituting back into the \( y_1 \) equation:
\[y_1 = \frac{2}{\sqrt{x} + \frac{1}{\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \left(1 - \frac{1}{x} \right)\]

Instead of further simplifying the above, we use an alternate method for substitution.

Alternate Simplification of the Original Expression:
\[y = \log \left( \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 \right) = \log \left( x + \frac{1}{x} + 2 \right) = \log \left( \frac{x^2 + 1 + 2x}{x} \right) = \log \left( \frac{(x + 1)^2}{x} \right)\]Using logarithmic properties:\[y = \log \left( \frac{(x + 1)^2}{x} \right) = \log (x + 1)^2 - \log x = 2 \log(x + 1) - \log x\]

4. Calculate the first derivative \( y_1 \):
\[y_1 = \frac{d}{dx} (2 \log(x + 1) - \log x) = 2 \cdot \frac{1}{x + 1} - \frac{1}{x}\]So, \( y_1 = \frac{2}{x + 1} - \frac{1}{x} \).

5. Calculate the second derivative \( y_2 \):
\[y_2 = \frac{d}{dx} \left( \frac{2}{x + 1} - \frac{1}{x} \right) = \frac{d}{dx} (2(x + 1)^{-1} - x^{-1}) = -2(x + 1)^{-2} - (-1)x^{-2} = -\frac{2}{(x + 1)^2} + \frac{1}{x^2}\]So, \( y_2 = -\frac{2}{(x + 1)^2} + \frac{1}{x^2} \).

6. Substitute \( y_1 \) and \( y_2 \) into the expression to be proved:
We aim to prove:\[x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2\]Substituting the calculated derivatives:

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

Simplify each term:

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 the simplified terms:

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

Combine the terms:

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

Conclusion:
The expression is proven to be equal to 2:\[x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2\]