Logarithmic Differentiation for $f(x)^{g(x)}$

1. The Differentiation Dilemma: Variable Base and Exponent

So far, you've mastered two key rules for functions with exponents:

1. The Power Rule: For $\frac{d}{dx}(x^n)$, where the base is a variable and the exponent is a constant, we get $n{x}^{n-1}$.
2. The Exponential Rule: For $\frac{d}{dx}(a^x)$, where the base is a constant and the exponent is a variable, we get $a^x\ln a$.

But what happens when we encounter a function where both the base and the exponent are variables, like $y=x^x$?

• We can't use the Power Rule because the exponent isn't a constant.
• We can't use the Exponential Rule because the base isn't a constant.

We need a new strategy. This is where logarithmic differentiation comes in. It's a clever technique that uses the properties of logarithms to transform the problem into something we can solve.

The following video offers a clear explanation of this exact problem.

2. The Technique: A Step-by-Step Guide

Logarithmic differentiation elegantly solves our dilemma by converting the problematic exponentiation into a product, which we can then handle using the product rule. The process always follows these four steps.

Guideline for Logarithmic Differentiation

1. Given $y=[f(x){]}^{g(x)}$, take the natural logarithm ($\ln$) of both sides: $\ln y=\ln ([f(x){]}^{g(x)})$.
2. Use the logarithm property $\ln (a^b)=b\ln (a)$ to bring the exponent down: $\ln y=g(x)\cdot \ln (f(x))$.
3. Differentiate both sides implicitly with respect to $x$. The left side will become $\frac{1}{y}\frac{dy}{dx}$, and the right side will usually require the product rule.
4. Solve for $\frac{dy}{dx}$ by multiplying the right side by $y$, and then substitute the original expression for $y$ back into the equation.

Let's apply this process to the classic example, $y=x^x$.

3. Applying the Technique: Worked Examples

We will now walk through a few examples, starting with the most fundamental one and gradually increasing in complexity.

Example 1: The Classic y = x^x

Let's differentiate $y=x^x$ using our four-step process.

1. Take $\ln$ of both sides:
   $\ln (y)=\ln (x^x)$
2. Use log properties:
   $\ln (y)=x\ln (x)$
3. Differentiate implicitly: We use the product rule on the right side.
   $\frac{d}{dx}(\ln y)=\frac{d}{dx}(x\ln x)$
   $\frac{1}{y}\frac{dy}{dx}=(1)\cdot \ln (x)+x\cdot \left(\frac{1}{x}\right)$
   $\frac{1}{y}\frac{dy}{dx}=\ln (x)+1$
4. Solve for $\frac{dy}{dx}$ and substitute:
   $\frac{dy}{dx}=y(\ln (x)+1)$
   $\frac{dy}{dx}=x^x(\ln (x)+1)$

That's our derivative. The following resources will help you solidify this fundamental example.

Example 2: Incorporating Trigonometric Functions and the Product Rule

Let's try a problem that you're likely to see on an exam, like $y=x^{\sin x}$. The process remains the same, but the differentiation step becomes more interesting.

Watch how this is solved in the same video from The Organic Chemistry Tutor.

Example 3: Combining with the Chain Rule

What if the function itself is more complex? Consider $y=(\ln x{)}^x$. Here, we'll need the chain rule when differentiating the logarithmic part.

1. Take $\ln$: $\ln y=\ln ((\ln x{)}^x)$
2. Use log properties: $\ln y=x\cdot \ln (\ln x)$
3. Differentiate implicitly: The right side requires the product rule.
   $\frac{1}{y}\frac{dy}{dx}=(1)\cdot \ln (\ln x)+x\cdot \frac{d}{dx}(\ln (\ln x))$
   To find $\frac{d}{dx}(\ln (\ln x))$, we use the chain rule. The derivative of $\ln (u)$ is $\frac{u^{\prime }}{u}$. Here, $u=\ln x$, so $u^{\prime }=\frac{1}{x}$.
   $\frac{d}{dx}(\ln (\ln x))=\frac{1/x}{\ln x}=\frac{1}{x\ln x}$
   Substituting this back:
   $\frac{1}{y}\frac{dy}{dx}=\ln (\ln x)+x\cdot \left(\frac{1}{x\ln x}\right)=\ln (\ln x)+\frac{1}{\ln x}$
4. Solve and substitute:
   $\frac{dy}{dx}=y\left(\ln (\ln x)+\frac{1}{\ln x}\right)=(\ln x{)}^x\left(\ln (\ln x)+\frac{1}{\ln x}\right)$

This example is a great test of your skills. The following video walks through it clearly.

A Note on Complex Products and Quotients

While the primary use of logarithmic differentiation is for $f(x{)}^{g(x)}$ functions, it can also be a powerful tool for simplifying the differentiation of functions with many products, quotients, and roots. For a function like $y=\frac{\sqrt{x+1}}{(x-2{)}^3(x+4{)}^5}$, applying the quotient and product rules would be tedious. Taking the natural log first can turn this into a simple sum and difference of derivatives. This is a useful trick to remember for MHT-CET to save time.

Conclusion

In this lesson, we've unlocked the method for differentiating functions of the form $y=[f(x){]}^{g(x)}$, a case where standard rules do not apply. By using the properties of logarithms, you can transform these complex exponential forms into manageable products, which can then be differentiated implicitly.

Key Takeaways:

• When to Use: Apply logarithmic differentiation when you have a function with a variable in both the base and the exponent.
• The Four-Step Process:
  1. Take the natural log ($\ln$) of both sides.
  2. Use log properties to bring the exponent down as a multiplier.
  3. Differentiate both sides implicitly with respect to $x$.
  4. Solve for $\frac{dy}{dx}$ and substitute the original function back in for $y$.
• Core Idea: This technique converts a variable^variable problem into a product rule problem, which you already know how to solve.

As you continue your journey through calculus, our next step will be to explore higher-order derivatives. This involves the straightforward but important process of differentiating a function multiple times to find its second derivative ($f^{\prime \prime }(x)$), third derivative, and so on. This concept is foundational for analyzing a function's curvature and for solving optimization problems.