Question

(a) If \( y = (\log x)^2 \), prove that \( x^2 y'' + x y' = 2 \).

Hint:

For differentiating composite functions, use the chain rule and carefully apply product and quotient rules for higher derivatives.

Answer 1

Step 1: {Find the first derivative of \( y \)}
Given \( y = (\log x)^2 \), we differentiate with respect to \( x \): \[ y' = 2 \log x \cdot \frac{1}{x} = \frac{2 \log x}{x}. \] Step 2: {Find the second derivative of \( y \)}
Now, differentiate \( y' = \frac{2 \log x}{x} \) with respect to \( x \) again: \[ y'' = \frac{d}{dx}\left( \frac{2 \log x}{x} \right) = \frac{2 \cdot \left( \frac{1}{x} \cdot x - \log x \right)}{x^2} = \frac{2 - 2 \log x}{x^2}. \] Step 3: {Substitute \( y' \) and \( y'' \) into the given expression}
We now substitute \( y' = \frac{2 \log x}{x} \) and \( y'' = \frac{2 - 2 \log x}{x^2} \) into the expression \( x^2 y'' + x y' \): \[ x^2 y'' + x y' = x^2 \cdot \frac{2 - 2 \log x}{x^2} + x \cdot \frac{2 \log x}{x}. \] Simplify: \[ x^2 y'' + x y' = 2 - 2 \log x + 2 \log x = 2. \] Conclusion: We have shown that \( x^2 y'' + x y' = 2 \).