Practice Maths

Differentiating Composite Log Functions and Applications

Key Terms

Chain rule
d/dx[f(g(x))] = f′(g(x)) × g′(x); differentiate the outer function, keep the inner, then multiply by the derivative of the inner.
Log of composite
d/dx[ln(g(x))] = g′(x) / g(x).
Log laws simplification
Expand ln(product), ln(quotient), or ln(power) using log laws BEFORE differentiating to reduce complexity.
Stationary points
Set f′(x) = 0 and solve; use f′′(x) to classify (positive = minimum, negative = maximum).
Tangent line
At (a, f(a)): gradient = f′(a); tangent equation: y − f(a) = f′(a)(x − a).
Applications
Logarithmic functions model many real-world phenomena; always check domain restrictions when differentiating.
Chain Rule for Logarithms
d/dx[ln(ax + b)] = a/(ax + b)
d/dx[ln(x² + 1)] = 2x/(x² + 1)

Simplify with Log Laws First:
d/dx[ln(x²√(x+1))] = d/dx[2 ln x + ½ ln(x+1)] = 2/x + 1/(2(x+1))

Applications:
Tangents and normals: use dy/dx at the point
Stationary points: solve dy/dx = 0

Exponential Decay:
If N = N0e−kt, then dN/dt = −kN
Worked Example 1: Differentiate y = ln(3x + 1).

Let f(x) = 3x + 1, so f′(x) = 3.
dy/dx = f′(x)/f(x) = 3/(3x + 1)
Worked Example 2: Find the equation of the tangent to y = ln(2x) at x = 1.

At x = 1: y = ln(2) ≈ 0.693
dy/dx = 2/(2x) = 1/x; at x = 1: gradient = 1
Tangent: y − ln 2 = 1(x − 1)
y = x − 1 + ln 2
Hot Tip: When you see a product or quotient inside a ln( ), always try expanding with log laws before differentiating. For example, ln(x2/(x+1)) = 2 ln x − ln(x+1), and differentiating term-by-term is far easier than applying the chain rule to the entire expression.

Composite Log Functions: Building on the Chain Rule

In the previous lesson, we established d/dx[ln(x)] = 1/x and the chain rule result d/dx[ln(f(x))] = f′(x)/f(x). In this lesson we deepen that skill, applying it to more complex composite functions and real-world rates of change problems.

The most common composite log forms you will encounter are: linear arguments like ln(ax + b), quadratic arguments like ln(x² + c), and expressions where log laws allow simplification before differentiation. Recognising which approach to use is a key exam skill.

Expanding with Log Laws Before Differentiating

The log laws — ln(ab) = ln a + ln b, ln(a/b) = ln a − ln b, ln(an) = n ln a — are not just algebraic tools. In calculus, they are differentiation shortcuts. When the argument of a logarithm is a product, quotient, or power, expanding it first turns a chain-rule problem into a simple sum of basic derivatives.

Example: Differentiate y = ln(x²/(x+1)). Using log laws: y = ln(x²) − ln(x+1) = 2 ln x − ln(x+1). Then dy/dx = 2/x − 1/(x+1). This is much cleaner than trying to differentiate the quotient directly under the ln sign.

Example: Differentiate y = ln((x+1)2). Using the power law: y = 2 ln(x+1). Then dy/dx = 2/(x+1). Direct chain rule would give 2(x+1)/(x+1)² = 2/(x+1) — same answer but more steps.

Finding Tangents and Normals to Log Curves

Tangent and normal problems require three steps: (1) find dy/dx; (2) substitute the given x-value to get the gradient; (3) use y − y1 = m(x − x1). For the normal, use gradient mnormal = −1/mtangent.

Always check whether the x-value is in the domain before substituting. For y = ln(2x − 3), you need x > 3/2. Substituting x = 1 would be invalid.

Stationary Points of Log Functions

Setting dy/dx = 0 for log functions often leads to equations like ln(x) = c or rational equations from the derivative. For y = x² − ln(x), dy/dx = 2x − 1/x = 0, which gives 2x² = 1, so x = 1/√2 (taking the positive root, since x > 0 in the domain).

Exponential Decay and Logarithmic Rates

Many real-world processes follow the model N = N0e−kt, where N0 is the initial value and k > 0 is the decay constant. Differentiating gives dN/dt = −kN0e−kt = −kN. This remarkable property — the rate of change is proportional to the current value — is the defining characteristic of exponential decay.

To find the half-life (time for N to halve): set N = N0/2, giving e−kt = 1/2, so −kt = −ln 2, and t = ln(2)/k. This formula is worth remembering.

Strategy for log differentiation: (1) Check if log laws apply to simplify first. (2) If not, apply d/dx[ln(f(x))] = f′(x)/f(x) directly. (3) Factorise or simplify the result if needed. (4) State the domain.

Mastery Practice

  1. Differentiate y = ln(3x + 1).
  2. Differentiate y = ln((x + 1)²).
  3. Differentiate y = ln(x²/(x + 1)). Use log laws first.
  4. Differentiate y = ln(ex + 1).
  5. Find the equation of the tangent to y = ln(2x) at x = 1.
  6. Find the stationary points of y = x² − ln(x) and state their nature.
  7. A population decays according to N = 500e−0.1t. Find dN/dt and calculate the rate of change when t = 10.
  8. Differentiate y = ln(√(x² + 4)). Simplify your answer.
  9. Find the normal to y = ln(x² + 1) at x = 1. State both gradient and equation.
  10. A radioactive substance decays: N = N0e−kt. After 20 years, only 60% remains. Find k (to 4 d.p.) and the half-life of the substance.
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