Before we begin, let's review some key properties of logarithms that will be useful:
Logarithm Properties:
We can find this derivative using implicit differentiation. This approach is not easy to compute using the limit definition, so we use a clever alternative method.
Theorem: For \(x > 0\),
\[\frac{d}{dx}(\ln x) = \frac{1}{x}\]Proof using Implicit Differentiation:
Starting with \(y = \ln x\), we know that this is equivalent to:
\[x = e^y\]Step 1: Take the derivative with respect to \(x\) on both sides:
\[\frac{d}{dx}(x) = \frac{d}{dx}(e^y)\] \[1 = e^y \cdot \frac{dy}{dx}\]Step 2: Solve for \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}\]Therefore:
\[\frac{d}{dx}(\ln x) = \frac{1}{x}\]General Formula: For \(x \neq 0\),
\[\frac{d}{dx}(\ln|x|) = \frac{1}{x}\]This formula works for both positive and negative values of \(x\):
Chain Rule Formula: If \(u = u(x)\) and \(u > 0\), then:
\[\frac{d}{dx}(\ln u) = \frac{1}{u} \cdot \frac{du}{dx}\]Solution:
Method 1: Simplify first using logarithm properties:
\[y = \ln(e^x) = x\]Therefore:
\[\frac{dy}{dx} = 1\]Method 2: Use the chain rule directly:
\[\frac{dy}{dx} = \frac{d}{dx}[\ln(e^x)] = \frac{1}{e^x} \cdot \frac{d}{dx}(e^x) = \frac{1}{e^x} \cdot e^x = 1\]Solution:
Method 1: Simplify using logarithm properties first:
\[y = \ln(\sin^4 x) = 4\ln(\sin x)\]Now differentiate:
\[\frac{dy}{dx} = 4 \cdot \frac{d}{dx}[\ln(\sin x)]\] \[= 4 \cdot \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x)\] \[= 4 \cdot \frac{1}{\sin x} \cdot \cos x\] \[= 4\cot x\]Method 2: Use chain rule directly (without simplifying first):
\[\frac{dy}{dx} = \frac{1}{\sin^4 x} \cdot \frac{d}{dx}(\sin^4 x)\] \[= \frac{1}{\sin^4 x} \cdot 4\sin^3 x \cdot \cos x\] \[= \frac{4\cos x}{\sin x} = 4\cot x\]Solution:
This is a complex expression, so it's best to simplify using logarithm properties first:
\[y = \ln\left[\frac{x^4 - 3x^2 + 1}{(x+5)(x-3)}\right]\] \[= \ln(x^4 - 3x^2 + 1) - \ln[(x+5)(x-3)]\] \[= \ln(x^4 - 3x^2 + 1) - [\ln(x+5) + \ln(x-3)]\] \[= \ln(x^4 - 3x^2 + 1) - \ln(x+5) - \ln(x-3)\]Now differentiate term by term:
\[y' = \frac{d}{dx}[\ln(x^4 - 3x^2 + 1)] - \frac{d}{dx}[\ln(x+5)] - \frac{d}{dx}[\ln(x-3)]\] \[= \frac{1}{x^4 - 3x^2 + 1} \cdot (4x^3 - 6x) - \frac{1}{x+5} - \frac{1}{x-3}\] \[= \frac{4x^3 - 6x}{x^4 - 3x^2 + 1} - \frac{1}{x+5} - \frac{1}{x-3}\]Important Limitation of Power Rule: The power rule \(\frac{d}{dx}(x^n) = nx^{n-1}\) applies only when the power is a constant.
For functions where both the base and exponent involve variables, we need a different technique: logarithmic differentiation.
Given \(y = [f(x)]^{g(x)}\), such as \(y = x^{2x}\), \(y = x^{\sin x}\), \(y = 7^x\), or \(y = x^x\):
Solution:
Step 1: Take \(\ln\) of both sides:
\[\ln y = \ln(x^{2x}) = 2x \cdot \ln x\]Step 2: Take \(\frac{d}{dx}\) of both sides:
\[\frac{d}{dx}(\ln y) = \frac{d}{dx}(2x \cdot \ln x)\]Left side (chain rule):
\[\frac{1}{y} \cdot \frac{dy}{dx}\]Right side (product rule):
\[\frac{d}{dx}(2x \cdot \ln x) = 2 \cdot \ln x + 2x \cdot \frac{1}{x} = 2\ln x + 2\]So we have:
\[\frac{1}{y} \cdot y' = 2\ln x + 2\]Step 3: Multiply both sides by \(y\):
\[y' = y(2\ln x + 2) = x^{2x}(2\ln x + 2)\]Solution:
Step 1: Take \(\ln\) of both sides:
\[\ln y = \ln(x^{\sin x}) = \sin x \cdot \ln x\]Step 2: Take \(\frac{d}{dx}\) of both sides:
\[\frac{1}{y} \cdot y' = \frac{d}{dx}(\sin x \cdot \ln x)\]Using the product rule on the right side:
\[\frac{1}{y} \cdot y' = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} = \cos x \cdot \ln x + \frac{\sin x}{x}\]Step 3: Multiply both sides by \(y\):
\[y' = y\left[\cos x \cdot \ln x + \frac{\sin x}{x}\right] = x^{\sin x}\left[\cos x \cdot \ln x + \frac{\sin x}{x}\right]\]Solution:
We use logarithmic differentiation:
Step 1: Take \(\ln\) of both sides:
\[\ln y = \ln(b^x) = x \cdot \ln b\]Step 2: Take \(\frac{d}{dx}\) of both sides:
\[\frac{1}{y} \cdot y' = \frac{d}{dx}(x \cdot \ln b) = \ln b\](Note: \(\ln b\) is a constant since \(b\) is a constant)
Step 3: Multiply both sides by \(y\):
\[y' = y \cdot \ln b = b^x \cdot \ln b\]Theorem: For any constant \(b > 0\),
\[\frac{d}{dx}(b^x) = b^x \cdot \ln b\]Special cases:
Solution:
We use implicit differentiation. Since \(y = \log_b x\), we have:
\[x = b^y\]Taking \(\frac{d}{dx}\) of both sides:
\[\frac{d}{dx}(x) = \frac{d}{dx}(b^y)\] \[1 = b^y \cdot \ln b \cdot \frac{dy}{dx}\]Solving for \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{1}{b^y \cdot \ln b} = \frac{1}{x \cdot \ln b}\](since \(b^y = x\))
Theorem: For any constant \(b > 0\), \(b \neq 1\),
\[\frac{d}{dx}(\log_b x) = \frac{1}{x \ln b}\]Special case when \(b = e\):
\[\frac{d}{dx}(\ln x) = \frac{1}{x \ln e} = \frac{1}{x \cdot 1} = \frac{1}{x}\]| Function | Derivative | Notes |
|---|---|---|
| \(\ln x\) | \(\frac{1}{x}\) | \(x > 0\) |
| \(\ln|x|\) | \(\frac{1}{x}\) | \(x \neq 0\) |
| \(\ln u\) | \(\frac{1}{u} \cdot \frac{du}{dx}\) | Chain rule, \(u > 0\) |
| \(\log_b x\) | \(\frac{1}{x \ln b}\) | \(b > 0\), \(b \neq 1\) |
| \(e^x\) | \(e^x\) | Special exponential |
| \(b^x\) | \(b^x \ln b\) | \(b > 0\) |
Key Takeaway: Logarithmic differentiation is a powerful technique for finding derivatives of functions that have the form \(y = [f(x)]^{g(x)}\), where both the base and exponent are functions of \(x\). The key steps are: