Calculating the Second Derivative of \(\rm \left(\frac{1}{x}\right)^x \) at \(\rm x=e\)

The problem asks us to find the value of \(\rm e^e\left(\frac{d^2 y}{d x^2}\right)_{x=e}\) for the function \(\rm y = \left(\frac{1}{x}\right)^x\).

First, let's rewrite the function \(\rm y\) in a more convenient form:

\(\rm y = \left(\frac{1}{x}\right)^x = (x^{-1})^x = x^{-x}\)

To find the derivatives of \(\rm y = x^{-x}\), we will use logarithmic differentiation.

Step 1: Apply Logarithmic Differentiation

Take the natural logarithm of both sides of the equation \(\rm y = x^{-x}\):

\(\rm \ln y = \ln(x^{-x})\)

Using the logarithm property \(\rm \ln(a^b) = b \ln a\), we get:

\(\rm \ln y = -x \ln x\)

Step 2: Differentiate Implicitly to Find the First Derivative

Now, differentiate both sides with respect to \(\rm x\). On the left side, we use the chain rule. On the right side, we use the product rule \(\rm \frac{d}{dx}(uv) = u'\ v + u\ v'\) with \(\rm u = -x\) and \(\rm v = \ln x\).

\(\rm \frac{d}{dx}(\ln y) = \frac{d}{dx}(-x \ln x)\)

\(\rm \frac{1}{y} \frac{dy}{dx} = - \left( \frac{d}{dx}(x) \cdot \ln x + x \cdot \frac{d}{dx}(\ln x) \right)\)

\(\rm \frac{1}{y} \frac{dy}{dx} = - \left( 1 \cdot \ln x + x \cdot \frac{1}{x} \right)\)

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

Now, solve for \(\rm \frac{dy}{dx}\):

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

Substitute back \(\rm y = x^{-x}\):

\(\rm \frac{dy}{dx} = -x^{-x}(\ln x + 1)\)

Step 3: Differentiate the First Derivative to Find the Second Derivative

We need to differentiate \(\rm \frac{dy}{dx} = -x^{-x}(\ln x + 1)\) with respect to \(\rm x\). We will use the product rule again, with \(\rm u = -x^{-x}\) and \(\rm v = (\ln x + 1)\).

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

\(\rm \frac{d^2 y}{dx^2} = - \left[ \frac{d}{dx}(x^{-x}) \cdot (\ln x + 1) + x^{-x} \cdot \frac{d}{dx}(\ln x + 1) \right]\)

We need the derivative of \(\rm x^{-x}\). From Step 2, we found that \(\rm \frac{1}{y} \frac{dy}{dx} = -(\ln x + 1)\) where \(\rm y = x^{-x}\). So, \(\rm \frac{d}{dx}(x^{-x}) = \frac{dy}{dx} = -x^{-x}(\ln x + 1)\).

Also, \(\rm \frac{d}{dx}(\ln x + 1) = \frac{1}{x} + 0 = \frac{1}{x}\).

Substitute these into the expression for \(\rm \frac{d^2 y}{dx^2}\):

\(\rm \frac{d^2 y}{dx^2} = - \left[ (-x^{-x}(\ln x + 1)) \cdot (\ln x + 1) + x^{-x} \cdot \frac{1}{x} \right]\)

\(\rm \frac{d^2 y}{dx^2} = - \left[ -x^{-x}(\ln x + 1)^2 + x^{-x} x^{-1} \right]\)

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

We can factor out \(\rm x^{-x}\):

\(\rm \frac{d^2 y}{dx^2} = x^{-x} \left[ (\ln x + 1)^2 - x^{-1} \right]\)

\(\rm \frac{d^2 y}{dx^2} = x^{-x} \left[ (\ln x + 1)^2 - \frac{1}{x} \right]\)

Step 4: Evaluate the Second Derivative at \(\rm x=e\)

Now, we need to find the value of \(\rm \frac{d^2 y}{dx^2}\) when \(\rm x = e\). Recall that \(\rm \ln e = 1\).

\(\rm \left(\frac{d^2 y}{dx^2}\right)_{x=e} = e^{-e} \left[ (\ln e + 1)^2 - \frac{1}{e} \right]\)

\(\rm \left(\frac{d^2 y}{dx^2}\right)_{x=e} = e^{-e} \left[ (1 + 1)^2 - \frac{1}{e} \right]\)

\(\rm \left(\frac{d^2 y}{dx^2}\right)_{x=e} = e^{-e} \left[ 2^2 - \frac{1}{e} \right]\)

\(\rm \left(\frac{d^2 y}{dx^2}\right)_{x=e} = e^{-e} \left[ 4 - \frac{1}{e} \right]\)

Step 5: Calculate the Final Required Value

The problem asks for \(\rm e^e\left(\frac{d^2 y}{d x^2}\right)_{x=e}\). Substitute the value we found in Step 4:

\(\rm e^e \left(\frac{d^2 y}{d x^2}\right)_{x=e} = e^e \cdot \left( e^{-e} \left[ 4 - \frac{1}{e} \right] \right)\)

\(\rm e^e \left(\frac{d^2 y}{d x^2}\right)_{x=e} = e^{e + (-e)} \left[ 4 - \frac{1}{e} \right]\)

\(\rm e^e \left(\frac{d^2 y}{d x^2}\right)_{x=e} = e^0 \left[ 4 - \frac{1}{e} \right]\)

Since \(\rm e^0 = 1\):

\(\rm e^e \left(\frac{d^2 y}{d x^2}\right)_{x=e} = 1 \cdot \left[ 4 - \frac{1}{e} \right] = 4 - \frac{1}{e}\)

Thus, the value of \(\rm e^e\left(\frac{d^2 y}{d x^2}\right)_{x=e}\) is \(\rm 4 - \frac{1}{e}\).

Revision Table: Key Concepts

This problem involved several key calculus concepts. Reviewing them is important for mastering differentiation.

• Logarithmic Differentiation: Used for functions of the form \(\rm f(x)^{g(x)}\). Takes the natural logarithm of both sides and then differentiates implicitly.
• Product Rule: \(\rm \frac{d}{dx}(uv) = u'v + uv'\). Used to differentiate products of functions.
• Chain Rule: Used when differentiating composite functions, like \(\rm \ln(y)\) with respect to \(\rm x\), which requires multiplying by \(\rm \frac{dy}{dx}\).
• Derivative of \(\rm \ln x\): \(\rm \frac{d}{dx}(\ln x) = \frac{1}{x}\).
• Derivative of \(\rm x^n\): \(\rm \frac{d}{dx}(x^n) = nx^{n-1}\).
• Evaluating Derivatives at a Point: Substituting a specific value of \(\rm x\) into the derivative expression.

Additional Information on Differentiation Techniques

Differentiating complex functions often requires combining multiple rules. For functions like \(\rm y = x^{-x}\), simple power rule or exponential rule does not apply directly because both the base and the exponent are functions of \(\rm x\).

Why Logarithmic Differentiation?

Logarithmic differentiation simplifies the process for functions like \(\rm y = f(x)^{g(x)}\). By taking the logarithm, the exponent comes down as a multiplier, turning the power into a product, which is easier to differentiate using the product rule:

\(\rm \ln y = \ln(f(x)^{g(x)}) = g(x) \ln(f(x))\)

Then, differentiating with respect to \(\rm x\) gives:

\(\rm \frac{1}{y}\frac{dy}{dx} = g'(x)\ln(f(x)) + g(x)\frac{f'(x)}{f(x)}\)

\(\rm \frac{dy}{dx} = y \left( g'(x)\ln(f(x)) + g(x)\frac{f'(x)}{f(x)} \right)\)

\(\rm \frac{dy}{dx} = f(x)^{g(x)} \left( g'(x)\ln(f(x)) + g(x)\frac{f'(x)}{f(x)} \right)\)

This general formula could also be used, but differentiating step-by-step as shown in the solution is often less prone to errors for specific problems.