Apply the chain rule: d/dx[ln(g(x))] = g'(x)/g(x). Here g(x) = x², so g'(x) = 2x. Therefore d/dx[ln(x²)] = 2x/x² = 2/x. The common error is forgetting the chain rule and writing 1/x². Note that ln(x²) = 2·ln(x), so you can also differentiate directly: d/dx[2·ln(x)] = 2·(1/x) = 2/x. Both approaches confirm the same answer.
Question 2 Multiple Choice
You need to differentiate f(x) = x^(sin x). A classmate says to use the power rule: f'(x) = sin(x) · x^(sin x - 1). Why is this wrong, and what is the correct approach?
AThe power rule applies here — the classmate's answer is correct
BThe power rule requires a constant exponent. Since sin(x) varies with x, you must use logarithmic differentiation: take ln of both sides, giving ln(f) = sin(x)·ln(x), then differentiate implicitly
CUse the chain rule directly: f'(x) = sin(x)·x^(sin x - 1) + x^(sin x)·cos(x)
DThe derivative doesn't exist because the exponent is not a polynomial
The power rule d/dx[x^n] = n·x^(n-1) requires n to be a constant. When the exponent itself is a function of x, the power rule fails. Logarithmic differentiation handles this: set y = x^(sin x), take ln of both sides: ln(y) = sin(x)·ln(x). Differentiate: (1/y)·y' = cos(x)·ln(x) + sin(x)/x. Solve: y' = x^(sin x)·[cos(x)·ln(x) + sin(x)/x]. The classmate only accounted for the base varying, not the exponent.
Question 3 True / False
The derivative of ln(2x) is 1/x, not 1/(2x).
TTrue
FFalse
Answer: True
By the chain rule: d/dx[ln(2x)] = (d/dx[2x])/(2x) = 2/(2x) = 1/x. Equivalently, ln(2x) = ln(2) + ln(x), and since ln(2) is a constant its derivative is 0, leaving d/dx[ln(x)] = 1/x. This is a common trap: students expect the 2 to appear in the derivative, but the chain rule's numerator and denominator both contain the factor 2, which cancels.
Question 4 True / False
The formula d/dx[log_b(x)] = 1/(x·ln(b)) shows that natural log (base e) is the 'natural' base for calculus because ln(e) = 1 simplifies the formula to 1/x.
TTrue
FFalse
Answer: True
Yes — for any base b, d/dx[log_b(x)] = 1/(x·ln(b)) because log_b(x) = ln(x)/ln(b) by change of base, and ln(b) is a constant. When b = e, ln(e) = 1, so the formula reduces to 1/x with no extra constant. This is one of several reasons e is the natural base for calculus: it produces the cleanest derivatives and integrals. For base 10: d/dx[log₁₀(x)] = 1/(x·ln(10)) ≈ 1/(2.303x).
Question 5 Short Answer
Explain how d/dx[ln(x)] = 1/x is derived from the relationship between ln and e^x — without looking it up.
Think about your answer, then reveal below.
Model answer: Set y = ln(x), which means e^y = x. Differentiate both sides with respect to x using the chain rule on the left: e^y · (dy/dx) = 1. Solve for dy/dx: dy/dx = 1/e^y = 1/x (substituting back e^y = x).
This derivation uses only two facts: the definition of ln as the inverse of e^x, and the chain rule. It illustrates why inverse function differentiation is powerful — you derive new results from known ones without memorizing additional formulas. The same technique derives d/dx[arcsin(x)], d/dx[arctan(x)], and other inverse function derivatives.