4.2 Derivatives of Exponential & Logarithmic Functions

4.2 Derivatives of Exponential & Logarithmic Functions

Quick Answer: The derivative of the natural exponential function is $\frac{d}{dx}[e^x] = e^x$, and by the Chain Rule, $\frac{d}{dx}[e^{g(x)}] = e^{g(x)} \cdot g'(x)$. For base $a > 0$, $\frac{d}{dx}[a^x] = a^x \ln a$. The derivative of the natural logarithm is $\frac{d}{dx}[\ln|x|] = \frac{1}{x}$, and for composites, $\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}$. Functions with variable bases and variable exponents ($y = x^x$ or $y = [u(x)]^{v(x)}$) cannot be differentiated with the power rule or exponential rule; they require logarithmic differentiation.

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1. The Natural Exponential Function

The mathematical constant $e \approx 2.71828$ is defined uniquely such that the exponential function $f(x) = e^x$ has a tangent line slope of exactly $1$ at its $y$-intercept $(0, 1)$:

$$\lim_{h \to 0} \frac{e^h - 1}{h} = 1$$

Applying the limit definition of the derivative:

$$\frac{d}{dx}[e^x] = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h \to 0} \frac{e^x(e^h - 1)}{h} = e^x \cdot \lim_{h \to 0} \frac{e^h - 1}{h} = e^x \cdot 1 = e^x$$

The Chain Rule for Natural Exponentials

When the exponent is a differentiable function $u = g(x)$:

$$\frac{d}{dx}\left[ e^{g(x)} \right] = e^{g(x)} \cdot g'(x)$$

Worked Examples

• Example A: $\frac{d}{dx}[e^{-4x}] = e^{-4x} \cdot (-4) = -4e^{-4x}$
• Example B: $\frac{d}{dx}[e^{x^3 - 2x}] = e^{x^3 - 2x} \cdot (3x^2 - 2) = (3x^2 - 2)e^{x^3 - 2x}$
• Example C (Product Rule combination): Differentiate $f(x) = x^3 e^{2x}$. $$f'(x) = \frac{d}{dx}[x^3] \cdot e^{2x} + x^3 \cdot \frac{d}{dx}[e^{2x}] = 3x^2 e^{2x} + x^3(2e^{2x}) = x^2 e^{2x}(3 + 2x)$$

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2. General Base Exponential Functions ($a^x$)

For any constant base $a > 0$ with $a \neq 1$, we convert to base $e$ using the logarithmic identity $a^x = e^{\ln(a^x)} = e^{x \ln a}$. Differentiating via the Chain Rule:

$$\frac{d}{dx}[a^x] = \frac{d}{dx}\left[ e^{x \ln a} \right] = e^{x \ln a} \cdot \frac{d}{dx}[x \ln a] = a^x \cdot \ln a$$

General Chain Rule Formula for Base $a$

$$\frac{d}{dx}\left[ a^{g(x)} \right] = a^{g(x)} \cdot (\ln a) \cdot g'(x)$$

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3. The Natural Logarithm and Absolute Values

The natural logarithm $y = \ln x$ is the inverse of $x = e^y$. Differentiating implicitly with respect to $x$:

$$\ne^y \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$$

The Absolute Value Extension $\ln|x|$

In calculus, logarithms often appear with absolute values to accommodate negative domain arguments ($x \neq 0$):

• For $x > 0$: $\ln|x| = \ln(x) \implies \frac{d}{dx}[\ln x] = \frac{1}{x}$.
• For $x < 0$: $\ln|x| = \ln(-x) \implies \frac{d}{dx}[\ln(-x)] = \frac{1}{-x} \cdot (-1) = \frac{1}{x}$.

$$\frac{d}{dx}[\ln|x|] = \frac{1}{x} \quad (x \neq 0)$$

The Chain Rule for Logarithms ("Derivative Over Function")

When differentiating the logarithm of a composite function $u = g(x)$:

$$\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}$$

General Base Logarithms ($\log_a x$)

Using the change-of-base formula $\log_a x = \frac{\ln x}{\ln a}$:

$$\frac{d}{dx}[\log_a x] = \frac{1}{x \ln a} \qquad \text{and} \qquad \frac{d}{dx}[\log_a(g(x))] = \frac{g'(x)}{g(x) \ln a}$$

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4. Logarithmic Properties as Pre-Differentiation Tools

When faced with the derivative of a logarithm containing complicated products, quotients, or powers, always expand the logarithm using algebraic properties before differentiating.

                      Logarithmic Expansion Strategy

        Complex Logarithm: ln [ (x³ · √(x²+1)) / (2x+5)⁴ ]
                               │
                               ▼  Apply Log Laws
        3 ln(x) + (1/2) ln(x²+1) - 4 ln(2x+5)
                               │
                               ▼  Differentiate Term-by-Term (No massive quotient rule!)
        3/x + (1/2) · [2x / (x²+1)] - 4 · [2 / (2x+5)]

The Three Fundamental Log Laws

1. Product Law: $\ln(u \cdot v) = \ln u + \ln v$
2. Quotient Law: $\ln\left(\frac{u}{v}\right) = \ln u - \ln v$
3. Power Law: $\ln(u^k) = k \ln u$

Worked Comparison: Expanding vs. Blind Differentiation

Problem: Find $f'(x)$ for $f(x) = \ln\left( \frac{x^4 \sqrt{3x - 1}}{(x^2 + 2)^5} \right)$.

• Inefficient Method (Blind Chain Rule): Applying $\frac{g'(x)}{g(x)}$ directly requires applying the Quotient Rule to two composite product factors—a recipe for catastrophic algebra mistakes.
• Efficient Method (Pre-Expansion First):
  1. Expand using log properties: $$f(x) = \ln(x^4) + \ln\left((3x - 1)^{1/2}\right) - \ln\left((x^2 + 2)^5\right)$$ $$f(x) = 4\ln(x) + \frac{1}{2}\ln(3x - 1) - 5\ln(x^2 + 2)$$
  2. Differentiate term-by-term instantly: $$f'(x) = 4\left(\frac{1}{x}\right) + \frac{1}{2}\left(\frac{3}{3x - 1}\right) - 5\left(\frac{2x}{x^2 + 2}\right)$$ $$f'(x) = \frac{4}{x} + \frac{3}{2(3x - 1)} - \frac{10x}{x^2 + 2}$$

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5. Logarithmic Differentiation: Variable Base and Variable Exponent

Standard rules fail when BOTH the base and the exponent contain variables: $y = [u(x)]^{v(x)}$.

• Power Rule fails because the exponent is not a constant.
• Exponential Rule fails because the base is not a constant.

The 4-Step Logarithmic Differentiation Protocol

1. Take the natural logarithm of both sides: $\ln y = \ln\left([u(x)]^{v(x)}\right) = v(x) \ln(u(x))$.
2. Differentiate implicitly with respect to $x$: $$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[v(x) \ln(u(x))]$$
3. Isolate $\frac{dy}{dx}$ by multiplying by $y$: $$\frac{dy}{dx} = y \cdot \frac{d}{dx}[v(x) \ln(u(x))]$$
4. Substitute the original expression back for $y$ so the final answer is purely in terms of $x$.

Classic Example: Differentiating $y = x^x$ for $x > 0$

1. Take natural logs: $\ln y = \ln(x^x) = x \ln x$.
2. Differentiate implicitly using Product Rule: $$\frac{1}{y} \frac{dy}{dx} = (1)(\ln x) + x\left(\frac{1}{x}\right) = \ln x + 1$$
3. Multiply by $y$: $$\frac{dy}{dx} = y(\ln x + 1)$$
4. Back-substitute $y = x^x$: $$\frac{d}{dx}[x^x] = x^x(\ln x + 1)$$

Example 2: Differentiating $y = (\sin x)^x$ on $(0, \pi)$

1. $\ln y = x \ln(\sin x)$
2. $\frac{y'}{y} = (1)\ln(\sin x) + x \cdot \frac{\cos x}{\sin x} = \ln(\sin x) + x \cot x$
3. $y' = (\sin x)^x [\ln(\sin x) + x \cot x]$

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6. CLEP Traps and Exam Strategies

• Trap 1: Misapplying Power Rule to Exponentials. A common error is writing $\frac{d}{dx}[e^x] = x e^{x-1}$ or $\frac{d}{dx}[2^x] = x 2^{x-1}$. The Power Rule is strictly for $(\text{variable})^{\text{constant}}$, never $(\text{constant})^{\text{variable}}$.
• Trap 2: Forgetting $\ln a$ in Base $a$ Exponential and Logarithmic Derivatives. On multiple-choice questions, options omitting $\ln a$ or putting $\ln a$ in the numerator instead of denominator for logarithms are classic distractor traps.
• Trap 3: Forgetting to Back-Substitute $y$ in Logarithmic Differentiation. Do not leave your final answer as $\frac{y'}{y} = \dots$. Always multiply across by $y$ and replace $y$ with its original formula in terms of $x$.