Problem 71 - Calculus for Biology and Medicine

Question

Use logarithmic differentiation to find the first derivative of the given functions. $$ y=x^{x^{x}} $$

Step-by-Step Solution

Answer

The first derivative is \( \frac{dy}{dx} = x^{x^{x}} \left(x^{x} \ln(x) (1 + \ln(x)) + x^{x-1}\right) \).

Step 1: Apply Logarithm to Both Sides

Start with the function \( y = x^{x^{x}} \). Apply the natural logarithm to both sides: \( \ln(y) = \ln(x^{x^{x}}) \). The logarithm of a power allows us to bring the exponent down in front as a product, so this becomes: \( \ln(y) = x^{x} \ln(x) \).

Step 2: Differentiate Implicitly

Differentiate both sides of the equation with respect to \( x \). For the left side, use the derivative of \( \ln(y) \), which is \( \frac{1}{y} \cdot \frac{dy}{dx} \). For the right side, use the product rule where you differentiate \( x^{x} \ln(x) \) with respect to \( x \).

Step 3: Differentiate the Right Side

Apply the product rule to \( x^{x} \ln(x) \), considering \( u = x^{x} \) and \( v = \ln(x) \). Recall that \( u = x^{x} \) and take its derivative using logarithmic differentiation: \( \ln(u) = x \ln(x) \) implies \( \frac{du}{dx} = x^{x} (1 + \ln(x)) \).

Step 4: Combine Derivatives Using Product Rule

The product rule states \((uv)' = u'v + uv'\). Substitute the derivatives: \( \frac{d}{dx} (x^{x} \ln(x)) = \left(x^{x} (1 + \ln(x))\right) \ln(x) + x^{x} \cdot \frac{1}{x} \). Simplify the expression: \( (x^{x} \ln(x) (1 + \ln(x))) + x^{x-1} \).

Step 5: Solve for \( \frac{dy}{dx} \)

Now substitute back into our implicit differentiation equation: \( \frac{1}{y} \cdot \frac{dy}{dx} = x^{x} \ln(x) (1 + \ln(x)) + x^{x-1} \). Multiply through by \( y \): \( \frac{dy}{dx} = y \left( x^{x} \ln(x) (1 + \ln(x)) + x^{x-1} \right) \).

Step 6: Replace \( y \) with Original Function

Since \( y = x^{x^{x}} \), substitute back to express \( \frac{dy}{dx} \) in terms of \( x \): \( \frac{dy}{dx} = x^{x^{x}} \left(x^{x} \ln(x) (1 + \ln(x)) + x^{x-1}\right) \).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when it's difficult or impossible to solve an equation for one variable in terms of another explicitly. In simple terms, it helps us find derivatives when the relationship between variables is intertwined. For example, if we have a function expressed implicitly, like in the equation \( \ln(y) = x^{x} \ln(x) \), we apply differentiation to both sides with respect to \(x\).
When differentiating both sides:

• The left side \(\frac{d}{dx}[\ln(y)]\) becomes \(\frac{1}{y} \cdot \frac{dy}{dx}\) because we're differentiating with respect to \(x\), but \(y\) is a function of \(x\).
• The right side is dealt with using other rules, like the product rule, as it's more complex.

Implicit differentiation is powerful because it lets us find \(\frac{dy}{dx}\) without needing to solve for \(y\) first. You simply differentiate, simplify, and then solve for \(\frac{dy}{dx}\). This becomes very useful in more advanced calculus problems.

Product Rule

The product rule is another fundamental concept in calculus used to find the derivative of a product of two functions. If you have a function given by \( u(x) \cdot v(x) \), the product rule tells you that the derivative \((uv)'\) is given by:\[ \frac{d}{dx}[u \cdot v] = u'v + uv' \]In our exercise, the function \( x^{x} \ln(x) \) requires us to apply this rule. Here's how it works step-by-step:

• Choose \(u = x^{x}\) and \(v = \ln(x)\).
• Find the derivative of each, \(u'\) and \(v'\).
• For \(u = x^{x}\), its derivative \(u'\) is found with logarithmic differentiation, resulting in \(x^{x}(1+\ln(x))\).
• For \(v = \ln(x)\), the derivative \(v'\) is simple: \(\frac{1}{x}\).
• Plug these into the product rule formula: \(\left(x^{x} (1 + \ln(x))\right) \ln(x) + x^{x} \cdot \frac{1}{x}\).

The product rule ensures that we properly account for how both parts of the function change, providing a complete picture of the derivative.

Derivatives of Exponential Functions

Exponentials can be tricky, especially when the exponent itself is another function. The key to differentiating exponential functions like \(x^{x}\) is understanding the power and chain rules.

For an exponential function \(a^{u(x)}\), the derivative involves both the original function and the derivative of its exponent:\[ \frac{d}{dx}[a^{u(x)}] = a^{u(x)} \cdot \ln(a) \cdot u'(x) \]However, in our log-based problem with \(x^{x}\), there's a special twist:

• First, express \(x^{x}\) in logarithmic form: \(\ln(u) = x \ln(x)\).
• Differentiating gives: \(\frac{du}{dx} = x^{x}(1+\ln(x))\).

This is because you're deriving a function raised to another function, which requires extra care.
By expanding the use of logarithms, these derivatives become manageable, allowing powerful simplifications that are handy in solving complex calculus problems.