Differentiation of Logarithmic Functions

TL;DR

You'll learn how to find the derivative of functions involving natural logarithms and logarithms with other bases. The core idea is using the chain rule with specific derivative formulas for logarithmic expressions. Logarithmic differentiation is a clever technique for finding derivatives of complex functions by taking the logarithm first.

1. The Mental Model

Think of differentiating logarithmic functions as a special application of the chain rule. You're essentially peeling back layers: first, the logarithm, then whatever function is inside it. Logarithmic differentiation is like using a "logarithmic magnifying glass" to simplify complicated products, quotients, and powers before differentiating.

2. The Core Material

When you differentiate a logarithmic function, you're usually working with either the natural logarithm (base $e$) or a logarithm with a different base. The good news is there are specific formulas for each, and the chain rule handles the "inside" part.

The Natural Logarithm: $\ln(x)$

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The derivative of the natural logarithm of $x$ is quite simple:

If $y = \ln(x)$, then $\frac{dy}{dx} = \frac{1}{x}$.

However, you'll almost always encounter $\ln(u)$, where $u$ is some function of $x$. This is where the chain rule comes in.

If $y = \ln(u(x))$, then $\frac{dy}{dx} = \frac{1}{u(x)} \cdot u'(x)$.

You can also write this as $\frac{u'}{u}$. This means you take the derivative of the inside function ($u'$) and divide it by the original inside function ($u$).

Logarithms with Other Bases: $\log_a(x)$

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What if the base isn't $e$? For example, $y = \log_a(x)$, where $a$ is some constant (like $\log_{10}(x)$).

You'll need a conversion formula or just remember the specific derivative:

If $y = \log_a(x)$, then $\frac{dy}{dx} = \frac{1}{x \ln(a)}$.

Again, with the chain rule for $y = \log_a(u(x))$:

$\frac{dy}{dx} = \frac{1}{u(x) \ln(a)} \cdot u'(x)$.

Notice it's very similar to the natural logarithm, but you divide by $\ln(a)$ as well.

Logarithmic Differentiation

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This is a powerful technique for functions that are products, quotients, or powers of other functions, especially when they look really messy. The idea is to take the natural logarithm of both sides of the equation first, use logarithm properties to expand and simplify, and then differentiate implicitly.

Here's the general process:

graph TD
    A["Start with y = f(x)"] --> B["Take ln of both sides: ln(y) = ln(f(x))"]
    B --> C["Use log properties to expand/simplify ln(f(x))"]
    C --> D["Differentiate both sides with respect to x (implicitly)"]
    D --> E["Solve for dy/dx"]
    E --> F["Substitute y back in (y = f(x))"]
    F --> G["Final Answer"]

Logarithm Properties you'll use:
* $\ln(AB) = \ln(A) + \ln(B)$
* $\ln(\frac{A}{B}) = \ln(A) - \ln(B)$
* $\ln(A^B) = B \ln(A)$

By using these, you turn complex multiplications/divisions/powers into simpler sums/differences, which are much easier to differentiate.

Example: Basic Natural Logarithm

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Let's find the derivative of $y = \ln(x^2 + 3x)$.

Here, $u(x) = x^2 + 3x$.
First, find $u'(x) = \frac{d}{dx}(x^2 + 3x) = 2x + 3$.
Now apply the formula $\frac{dy}{dx} = \frac{u'}{u}$:
$\frac{dy}{dx} = \frac{2x + 3}{x^2 + 3x}$.

Example: Logarithm with Another Base

Find the derivative of $y = \log_2(\sin(x))$.

Here, $u(x) = \sin(x)$ and $a = 2$.
First, find $u'(x) = \frac{d}{dx}(\sin(x)) = \cos(x)$.
Now apply the formula $\frac{dy}{dx} = \frac{u'}{u \ln(a)}$:
$\frac{dy}{dx} = \frac{\cos(x)}{\sin(x) \ln(2)}$.
You could also write this as $\frac{\cot(x)}{\ln(2)}$.

3. Worked Example

Let's use logarithmic differentiation to find the derivative of $y = \frac{(x^2+1)^3 \cdot e^{2x}}{\sqrt{x}}$.

1. Take the natural logarithm of both sides:
   $\ln(y) = \ln\left(\frac{(x^2+1)^3 \cdot e^{2x}}{\sqrt{x}}\right)$

2. Use logarithm properties to expand:
   $\ln(y) = \ln((x^2+1)^3) + \ln(e^{2x}) - \ln(\sqrt{x})$
   $\ln(y) = 3\ln(x^2+1) + 2x\ln(e) - \frac{1}{2}\ln(x)$
   Since $\ln(e) = 1$:
   $\ln(y) = 3\ln(x^2+1) + 2x - \frac{1}{2}\ln(x)$

3. Differentiate both sides implicitly with respect to $x$:
   The derivative of $\ln(y)$ is $\frac{1}{y} \frac{dy}{dx}$.
   The derivative of $3\ln(x^2+1)$ is $3 \cdot \frac{2x}{x^2+1} = \frac{6x}{x^2+1}$.
   The derivative of $2x$ is $2$.
   The derivative of $-\frac{1}{2}\ln(x)$ is $-\frac{1}{2} \cdot \frac{1}{x} = -\frac{1}{2x}$.

   So, we get:
   $\frac{1}{y} \frac{dy}{dx} = \frac{6x}{x^2+1} + 2 - \frac{1}{2x}$

4. Solve for $\frac{dy}{dx}$:
   Multiply both sides by $y$:
   $\frac{dy}{dx} = y \left( \frac{6x}{x^2+1} + 2 - \frac{1}{2x} \right)$

5. Substitute the original expression for $y$ back in:
   $\frac{dy}{dx} = \frac{(x^2+1)^3 \cdot e^{2x}}{\sqrt{x}} \left( \frac{6x}{x^2+1} + 2 - \frac{1}{2x} \right)$

That's your final derivative! Much easier than using the product, quotient, and chain rules repeatedly from the start.

4. Key Takeaways

• The derivative of $\ln(u)$ is $\frac{u'}{u}$. Remember to find the derivative of the "inside" function $u$.
• The derivative of $\log_a(u)$ is $\frac{u'}{u \ln(a)}$. It's similar to $\ln(u)$, but with an extra $\ln(a)$ in the denominator.
• Logarithm properties are crucial for simplifying expressions before differentiation, especially when dealing with complex products, quotients, or powers.
• Logarithmic differentiation involves taking the natural log of both sides, expanding, differentiating implicitly, and then solving for $\frac{dy}{dx}$.
• This technique is particularly useful for functions of the form $f(x)^{g(x)}$ (a function raised to the power of another function) because it brings the exponent down.
• Always substitute the original function back in for $y$ at the end of logarithmic differentiation.
• Don't forget the chain rule! It's implicitly used whenever $u(x)$ is more than just $x$.

5. Now Try It

Find the derivative of the function $y = x^{\sin(x)}$. This is a perfect candidate for logarithmic differentiation since it's a function raised to another function. When you're done, your solution should include $\ln(x)$ and $\cos(x)$ terms, as well as the original $x^{\sin(x)}$ in your final expression for $\frac{dy}{dx}$.