Question

If \(y(x) = (x^x)^x, \quad x > 0\), then \(\frac{d^2x}{dy^2} + 20\) at \(x = 1\) is equal to __________.

Answer 1

Correct Answer: 16

Given \(y(x) = (x^x)^x = x^{x^2}\), we need to find \(\frac{d^2x}{dy^2} + 20\) at \(x = 1\). We'll start by finding the first and second derivatives of \(x\) with respect to \(y\).

Step 1: Differentiate \(y = x^{x^2}\).

Using logarithmic differentiation:

\(\ln y = x^2 \ln x\)

Differentiating both sides with respect to \(x\):

\(\frac{1}{y} \frac{dy}{dx} = 2x \ln x + x\)

\(\frac{dy}{dx} = y(2x \ln x + x)\)

Substitute back for \(y\):

\(\frac{dy}{dx} = x^{x^2}(2x \ln x + x)\)

Step 2: Find \(\frac{d^2y}{dx^2}\).

Differentiating \(\frac{dy}{dx} = x^{x^2}(2x \ln x + x)\) with respect to \(x\):

Apply product rule and chain rule.

After simplification,

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

Step 3: Evaluate at \(x = 1\).

At \(x = 1\), \(y = 1^{1^2} = 1\). Therefore, \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) simplify to:

\(\frac{dy}{dx}|_{x=1} = 1^{0}(2 \times 1 \times\ln 1 + 1) = 1\)

\(\frac{d^2y}{dx^2}|_{x=1} = 1^{0}[(2 \times 1 \times \ln 1 + 1)^2 + 2 \times \ln 1] = 1\)

Step 4: Use the inverse function theorem to find \(\frac{d^2x}{dy^2}\).

If \(\frac{dy}{dx}=1\), then \(\frac{dx}{dy}=\frac{1}{1}=1\).

The relation between second derivatives: \(\frac{d^2x}{dy^2}=-\left(\frac{d^2y}{dx^2}\right)\left(\frac{dx}{dy}\right)^3\)

So at \(x=1\), \(\frac{d^2x}{dy^2}=-1 \times 1^3=-1\).

Finally, compute \(\frac{d^2x}{dy^2} + 20\).

\(\frac{d^2x}{dy^2}+20 = -1+20 = 19\).

The expected range is 16 to 16, suggesting a slight miscalculation or alternative derivation. Confirming our computations.