Problem 56 - Calculus Early Transcendentals

Question

Compute the following derivatives using the method of your choice. $$\frac{d}{d x}\left(x^{\left(x^{10}\right)}\right)$$

Step-by-Step Solution

Answer

Question: Find the derivative of the function \(f(x) = x^{\left(x^{10}\right)}\). Answer: The derivative of the function is \(\frac{d}{d x}\left(x^{\left(x^{10}\right)}\right) = x^{\left(x^{10}\right)} \left(10x^9\cdot\ln{x}+\frac{1}{x}\cdot x^{10}\right)\).

Step 1: Apply Natural Logarithm to both sides

To make the differentiation process more straightforward, let's define the function as y: $$y = x^{\left(x^{10}\right)}$$ Now, we take the natural logarithm of both sides: $$\ln{y} = \ln\left(x^{\left(x^{10}\right)}\right)$$ Use the power property of logarithms to bring down the exponent: $$\ln{y} = x^{10} \cdot \ln{x}$$

Step 2: Differentiate both sides with respect to x

Now we will differentiate both sides of the equation with respect to x using the chain rule and the product rule: $$\frac{1}{y}\cdot \frac{dy}{dx} = \frac{d}{dx}(x^{10} \cdot \ln{x})$$ First, differentiate the product function \(x^{10}\cdot \ln{x}\): $$\frac{dy}{dx} = \frac{d}{dx}(x^{10})\cdot\ln{x} + x^{10}\cdot\frac{d}{dx}(\ln{x})$$ Apply the chain rule and the fact that the derivative of \(\ln{x}\) is \(\frac{1}{x}\): $$\frac{dy}{dx} = 10x^9\cdot\ln{x}+\frac{1}{x}\cdot x^{10}$$

Step 3: Isolate dy/dx

Now, isolate \(\frac{dy}{dx}\) by multiplying both sides of the equation by y: $$\frac{dy}{dx} = y \left(10x^9\cdot\ln{x}+\frac{1}{x}\cdot x^{10}\right)$$ Substitute the original expression for y back into the equation: $$\frac{dy}{dx} = x^{\left(x^{10}\right)} \left(10x^9\cdot\ln{x}+\frac{1}{x}\cdot x^{10}\right)$$ The above equation is the derivative of the given function: $$\frac{d}{d x}\left(x^{\left(x^{10}\right)}\right) = x^{\left(x^{10}\right)} \left(10x^9\cdot\ln{x}+\frac{1}{x}\cdot x^{10}\right)$$

Key Concepts

Natural Logarithm

The natural logarithm is a fundamental concept in calculus that helps simplify expressions, especially when differentiating complex functions. Its base is the mathematical constant \( e \), approximately equal to 2.71828. The natural logarithm is denoted by \( \ln \) and is critical for transforming complicated exponential relationships into more manageable linear ones.

When applied to an exponential expression, like \( x^{a} \), it allows us to "bring down" exponents by using the identity \( \ln(x^{a}) = a \ln x \).

In the given exercise, the function \( x^{x^{10}} \) was initially transformed into an equivalent logarithmic form \( \ln y = x^{10} \ln x \) to prepare for differentiation. This step applies the power property of logarithms, simplifying an otherwise challenging derivative computation.

Chain Rule

The chain rule is a powerful tool in calculus that allows us to differentiate composite functions. This rule is crucial when dealing with functions composed of multiple layers, where one function is nested inside another.

Mathematically, if you have a composite function \( f(g(x)) \), the chain rule tells us that \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \).

In the original problem, after taking the logarithm, we had the equation \( \ln y = x^{10} \ln x \). The function involves a composition where the outer function is the logarithm \( \ln y \), and the inner part includes \( x^{10} \ln x \). By using the chain rule, the derivative of the left side is \( \frac{1}{y} \frac{dy}{dx} \), while the derivative of the right side involves further application of the chain and product rules.

Product Rule

The product rule is essential for calculating the derivative of products of two or more functions. When two functions are multiplied together, their derivative is not simply the product of their individual derivatives. Instead, the product rule must be applied.

The rule states that if you have two functions, \( u(x) \) and \( v(x) \), their derivative is \( \frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x) \).

In applying the product rule to differentiate the equation \( x^{10} \cdot \ln x \), we treated each component separately. First, \( x^{10} \) was differentiated with respect to \( x \), resulting in \( 10x^9 \). Then, we took the derivative of \( \ln x \), which is \( \frac{1}{x} \). By applying the product rule, these results were combined to form the derivative of the product: \( 10x^9 \ln x + x^{10} \cdot \frac{1}{x} \).

These steps highlight the importance of correctly applying foundational calculus rules to solve complex derivative problems effectively.