Exponential and Logarithmic Derivatives

1. Foundations: Index and Logarithmic Laws as Differentiation Tools

Before differentiating exponential and logarithmic functions, it is essential to be fluent in manipulating expressions using index and logarithmic laws. These laws are not merely algebraic tools — they actively simplify functions into forms that are far easier to differentiate, and they allow us to verify or rearrange results once derivatives have been found.

Key index laws used in this topic:

• Product rule (indices): a^m · aⁿ = a^m+n
• Quotient rule (indices): a^m / aⁿ = a^m−n
• Power law: (a^m)ⁿ = a^mn
• Negative index: a⁻ⁿ = 1/aⁿ

Key logarithm laws (for base e, i.e. natural logarithm ln):

• Log of a product: ln(ab) = ln a + ln b
• Log of a quotient: ln(a/b) = ln a − ln b
• Log of a power: ln(aⁿ) = n ln a
• Inverse relationship: e^{ln x} = x for x > 0, and ln(e^x) = x for every real x

These laws become especially powerful when applied before differentiating. For example, ln(x³) simplifies to 3 ln x, making the derivative immediately 3/x rather than requiring the chain rule on a composite expression. Similarly, expressions like e^{3x + 1} · e^x can be combined into e^{4x + 1} before differentiating.

Worked Example: Simplify ln(x²e^x) before differentiating.
First the domain: x²e^x > 0 for every x ≠ 0, so the expression is defined on x ≠ 0.
Using log laws: ln(x²e^x) = ln(x²) + ln(e^x) = 2 ln|x| + x (the modulus is what keeps the x < 0 branch; write 2 ln x + x only once you have restricted to x > 0)
Differentiating: d/dx[2 ln|x| + x] = 2/x + 1, valid for x ≠ 0
This is far more efficient than applying the product rule directly inside a logarithm.

2. Differentiating Exponential Functions: d/dx[e^(f(x))] = f′(x)e^(f(x))

The derivative of the natural exponential function is its defining property: e^x is its own derivative. When the exponent is itself a function of x — a composite form — the chain rule gives us the general result:

d/dx[e^f(x)] = f′(x) · e^f(x)

This result follows directly from the chain rule. Let u = f(x), so y = e^u. Then dy/dx = (dy/du) · (du/dx) = e^u · f′(x) = f′(x)e^f(x). The original function is preserved and simply multiplied by the derivative of the exponent.

Immediate results:

• d/dx[e^x] = e^x (the base case, where f(x) = x and f′(x) = 1)
• d/dx[e^3x] = 3e^3x
• d/dx[e^−2x] = −2e^−2x
• d/dx[e^x²] = 2xe^x²
• d/dx[e^{sin x}] = cos x · e^{sin x}

Worked Example 1 — Simple chain rule:
Find dy/dx for y = e^{4x − 1}.
Here f(x) = 4x − 1, so f′(x) = 4.
Therefore dy/dx = 4e^{4x − 1}.

Worked Example 2 — Combining with the product rule:
Find d/dx[x² e^3x].
Let u = x² and v = e^3x.
Then u′ = 2x and v′ = 3e^3x.
By the product rule: d/dx[x² e^3x] = u′v + uv′ = 2x · e^3x + x² · 3e^3x = e^3x(2x + 3x²) = xe^3x(2 + 3x).

Worked Example 3 — Quotient form:
Find d/dx[e^2x / x].
Using the quotient rule with u = e^2x, v = x:
d/dx[e^2x / x] = (2e^2x · x − e^2x · 1) / x² = e^2x(2x − 1) / x².

A common Band A extension is to simplify the result by factoring out the exponential term, which is always positive and therefore never cancels a sign or creates a zero — useful when finding stationary points.