Question

If \( y(x) = x^x, \, x > 0 \), then \( y''(2) - 2y'(2) \) is equal to:

• A. \( 4 (\log_e 2)^2 - 2 \)
• B. \( 8 \log_e 2 - 2 \)
• C. \( 4 (\log_e 2)^2 + 2 \)
• D. \( 4 \log_e 2 + 2 \)

Hint:

When working with functions like x^x, take the natural logarithm for simplification. This often helps to differentiate effectively using logarithmic properties.

Answer 1

The Correct Option is A

We are tasked with finding \(y''(2) - 2y'(2)\) for the function \(y(x) = x^x,\ x > 0\).

First, express the function in a form that makes differentiation easier:

Let \(y = x^x\). We can write this as:

\(y = e^{x \log_e x}\).

Now, differentiate \(y\) with respect to \(x\):

Find the first derivative, \(y'\):

Using the chain rule for differentiation, we have:

\(\frac{dy}{dx} = \frac{d}{dx}[e^{x \log_e x}] = e^{x \log_e x} \cdot \frac{d}{dx}(x \log_e x)\).

Now, find \(\frac{d}{dx}(x \log_e x)\):

\(= \log_e x + x \cdot \frac{1}{x} = \log_e x + 1\).

Therefore:

\(\frac{dy}{dx} = e^{x \log_e x} \cdot (\log_e x + 1) = x^x \cdot (\log_e x + 1)\).

Find the second derivative, \(y''\):

Differentiate \(y' = x^x (\log_e x + 1)\):

Apply the product rule:

\(\frac{d^2y}{dx^2} = \frac{d}{dx}[x^x] \cdot (\log_e x + 1) + x^x \cdot \frac{d}{dx}(\log_e x + 1)\).

We already found \(\frac{d}{dx}[x^x] = x^x (\log_e x + 1)\).

Now find \(\frac{d}{dx}(\log_e x + 1) = \frac{1}{x}\):

Substitute it back:

\(\frac{d^2y}{dx^2} = x^x (\log_e x + 1)^2 + x^x \cdot \frac{1}{x}\).

Simplify:

\(\frac{d^2y}{dx^2} = x^x (\log_e x + 1)^2 + x^{x-1}\).

Now compute \(y''(2)\) and \(y'(2)\):

\(y'(2) = 2^2 (\log_e 2 + 1) = 4 (\log_e 2 + 1)\).

Calculate \(y''(2)\):

\(y''(2) = 2^2 (\log_e 2 + 1)^2 + 2^{2-1}\).

Therefore:

\(y''(2) = 4 (\log_e 2 + 1)^2 + 2\).

Now expand \((\log_e 2 + 1)^2\):

\(= (\log_e 2)^2 + 2\log_e 2 + 1\).

So:

\(y''(2) = 4((\log_e 2)^2 + 2\log_e 2 + 1) + 2\).

\(= 4(\log_e 2)^2 + 8\log_e 2 + 4 + 2\).

\(= 4(\log_e 2)^2 + 8\log_e 2 + 6\).

Finally, calculate \(y''(2) - 2y'(2)\):

\(y''(2) - 2y'(2) = [4(\log_e 2)^2 + 8\log_e 2 + 6] - 2[4 (\log_e 2 + 1)]\).

\(= 4(\log_e 2)^2 + 8\log_e 2 + 6 - 8\log_e 2 - 8\).

Simplify:

\(= 4(\log_e 2)^2 - 2\).

Therefore, the result, \(y''(2) - 2y'(2)\), is \(4 (\log_e 2)^2 - 2\), which matches the given correct answer.