Question

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}(\ln x)^{x^{2}}$$

Short answer

Answer: The derivative of the function \((\ln x)^{x^2}\) with respect to \(x\) is \((\ln x)^{x^2} \left(2x\ln(\ln x) + \frac{x}{\ln x}\right)\).

Step-by-step solution

Write down the function

Consider the function: $$y = (\ln x)^{x^2}$$

Apply Logarithmic Differentiation

We will take the natural logarithm of both sides and differentiate implicitly with respect to \(x\). First, apply the natural logarithm: $$\ln y = x^2 \ln (\ln x)$$

Differentiate both sides with respect to x

Differentiate the equation with respect to \(x\), recalling that we'll need to use the chain rule on the right-hand side and use the fact that \(\frac{d}{dx}\ln y = \frac{1}{y}\frac{dy}{dx}\): $$\frac{1}{y}\frac{dy}{dx} = 2x\ln(\ln x) + x^2 \frac{1}{\ln x} \frac{d}{dx}\ln x$$

Simplify the expression

Simplify the expression by using the chain rule on the last term, since \(\frac{d}{dx}\ln x = \frac{1}{x}\): $$\frac{1}{y}\frac{dy}{dx} = 2x\ln(\ln x) + x^2 \frac{1}{\ln x} \cdot \frac{1}{x}$$ $$\frac{1}{y}\frac{dy}{dx} = 2x\ln(\ln x) + \frac{x}{\ln x}$$

Solve for the derivative

Now solve for \(\frac{dy}{dx}\) by multiplying both sides by \(y\) (recall that \(y = (\ln x)^{x^2}\)): $$\frac{dy}{dx} = y \left(2x\ln(\ln x) + \frac{x}{\ln x}\right) = (\ln x)^{x^2} \left(2x\ln(\ln x) + \frac{x}{\ln x}\right)$$ Finally, the derivative of the function with respect to \(x\) is: $$\frac{d}{dx}(\ln x)^{x^2} = (\ln x)^{x^2} \left(2x\ln(\ln x) + \frac{x}{\ln x}\right)$$

Key concepts

Implicit Differentiation

Implicit differentiation is a handy technique used in calculus when a function is not given in the standard form, which is 'explicitly' as y=f(x). Instead, when a function is given in an 'implicit' form involving x and y, like an equation that can't easily be solved for y, implicit differentiation allows us to find the derivative of y with respect to x without isolating y.

Here's how it works: we differentiate both sides of the equation with respect to x, treating y as a 'function of x'—even if we don't know what that function looks like. This often involves using the chain rule (more on that in a moment) to differentiate terms containing y. We eventually solve for dy/dx to get the derivative.

For instance, in our exercise, we applied logarithmic differentiation (which is a form of implicit differentiation) to the function \(y = (\ln x)^{x^2}\) by first taking the natural logarithm of both sides and then differentiating, treating y as a function of x. This provided us with a way to find the derivative of a complex function that would be difficult to differentiate otherwise.

Chain Rule

The chain rule is one of the quintessential tools in differentiation. It comes into play when we need to differentiate a function that is composed of other functions—a 'function of a function', so to speak. In essence, the chain rule tells us that to find the derivative of a composite function, we need to take the derivative of the outer function and multiply it by the derivative of the inner function.

An analogy often used is to imagine a machine within a machine. The rate at which the whole changes (the derivative) depends on the rate at which each individual part changes. Mathematically, if \(h(x) = f(g(x))\), then the derivative \(h'(x) = f'(g(x))\cdot g'(x)\).

In the given exercise, the chain rule was necessary when differentiating \(\ln(\ln x)\), as it is a composite function with \(\ln x\) nestled inside another logarithm. We first differentiated the outer function and then multiplied by the derivative of the inner function—exactly as the chain rule dictates.

Natural Logarithm

The natural logarithm, denoted as \(\ln x\), is the logarithm to the base \(e\), where \(e\) is an irrational and transcendental constant approximately equal to 2.71828. The natural logarithm is the inverse operation of raising \(e\) to a power, so if \(y = \ln x\), then we can say \(e^y = x\).

In calculus, the natural logarithm is critical because it has a derivative that is simple and elegant: \(\frac{d}{dx}\ln x = \frac{1}{x}\), for \(x > 0\). This property makes it particularly useful for integrating functions and solving differential equations. Logarithmic differentiation—a technique showcased in our exercise—involves taking the logarithm of a function to simplify the differentiation process, especially when dealing with products, quotients, or powers.

Exponential Derivatives

Exponential derivatives refer to the derivatives of functions that have a variable as an exponent, such as \(e^{x}\) or \(a^{x}\) where \(a\) is a constant. The derivative of \(e^{x}\) is particularly simple—it's just \(e^{x}\) itself. When it comes to differentiating other exponential functions like \(a^{x}\), we use the formula \(\frac{d}{dx}a^{x} = a^{x}\ln(a)\).

Exponential derivatives are also involved in implicit and logarithmic differentiation. For example, in our textbook solution, when we took the natural logarithm of both sides of the equation \(y = (\ln x)^{x^2}\), we effectively transformed an exponential function into a form where we could conveniently apply the properties of logarithms and the rules of differentiation to find the derivative.