The Natural Logarithm and Differentiating ln(x)

The Natural Logarithm: What It Is and Where It Comes From

The natural logarithm ln(x) is defined as log_e(x), where e ≈ 2.71828 is Euler’s number. While this might seem like an arbitrary base, ln(x) is in fact the most fundamental logarithm in calculus — not base 10. The reason becomes clear from its derivative.

Historically, the natural logarithm was discovered through the problem of finding the area under the hyperbola y = 1/t from t = 1 to t = x. This area turns out to equal ln(x). That is: ∫₁^x (1/t) dt = ln(x). This connection between 1/x and its antiderivative is precisely why the derivative of ln(x) is so clean.

The Derivative of ln(x)

The key result is: d/dx[ln(x)] = 1/x for x > 0. This is one of the most important derivatives you will use. It fills a crucial gap — the power rule ∫xⁿ dx = xⁿ⁺¹/(n+1) + c fails when n = −1. The natural logarithm provides the antiderivative of 1/x: ∫ 1/x dx = ln|x| + c. The absolute value handles negative values of x.

Using the chain rule, if y = ln(f(x)), then dy/dx = f′(x)/f(x). This is sometimes described as “the derivative of what’s inside, divided by what’s inside.” For example, d/dx[ln(x² + 4)] = 2x/(x² + 4).

Using Log Laws Before Differentiating

The log laws are powerful simplification tools. When you see a product, quotient, or power inside a logarithm, it is often faster to expand using log laws before differentiating, rather than applying the chain rule directly.

For example, to differentiate y = ln(x³), you could use the chain rule directly (getting 3x²/x³ = 3/x), or you could first write y = 3 ln(x) using the power law, giving dy/dx = 3/x immediately — much simpler.

Similarly, ln(x²√(x+1)) = 2 ln x + ½ ln(x+1). Differentiating term by term: 2/x + 1/(2(x+1)). This avoids a messy chain rule calculation on the combined expression.

Domain and Graphical Features of ln(x)

The function y = ln(x) has domain x > 0 only. It passes through (1, 0) since ln(1) = 0, and through (e, 1) since ln(e) = 1. As x → 0⁺, ln(x) → −∞ (vertical asymptote at x = 0). As x → ∞, ln(x) → ∞ but very slowly — much slower than any power of x.

The gradient function 1/x is always positive for x > 0, confirming that ln(x) is always increasing. The second derivative is −1/x² < 0, confirming the curve is always concave down. There are no stationary points.

Finding Tangent Lines and Stationary Points

Tangent line problems follow the standard process: find the gradient dy/dx at the given x-value, then use y − y₁ = m(x − x₁). For the function y = x ln(x), we need the product rule: dy/dx = ln(x) + x × (1/x) = ln(x) + 1. Setting this to zero: ln(x) = −1 ⇒ x = e⁻¹ = 1/e. At this point y = (1/e) ln(1/e) = (1/e)(−1) = −1/e. So the stationary point is (1/e, −1/e).