medium difficulty8 min

Logarithmic Differentiation

Logarithmic differentiation is a method where logarithms are taken on both sides before differentiating, especially for variable powers and complicated products or quotients.

In one line: Log first, expand neatly, differentiate, multiply back by y.

Practice This Concept

Understand it

Use this method when y has the form u(x)^v(x), or when y is a product and quotient of many factors. Taking log converts powers into products, products into sums, and quotients into differences, making differentiation manageable.

Example

If y = x^x, then log y = x log x. Differentiating gives (1/y)dy/dx = log x + 1, so dy/dx = x^x(log x + 1).

Formula and rule

If y = u(x)^v(x) with u(x) > 0, then log y = v(x)log u(x), and dy/dx = y[v'(x)log u(x) + v(x)u'(x)/u(x)].

Product and quotient forms use log(ab) = log a + log b and log(a/b) = log a - log b, with positive factors where real logarithms are used.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

Differentiate y = x² √(x + 1)/(x - 3), where factors are positive in the working interval.

1Taking logs: log y = 2log x + (1/2)log(x + 1) - log(x - 3).
2Differentiate: (1/y)dy/dx = 2/x + 1/[2(x + 1)] - 1/(x - 3).
3Therefore dy/dx = [x² √(x + 1)/(x - 3)][2/x + 1/(2x + 2) - 1/(x - 3)].

Method you can reuse

1Set the expression equal to y.
2Take log on both sides.
3Use logarithm laws to expand powers, products, and quotients.
4Differentiate both sides with respect to x.
5Solve for dy/dx by multiplying by y.
6Substitute the original expression for y.

Visual tip

No diagram is required; the method depends on algebraic transformation using logarithm laws.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

Differentiate y = (x² + 1)^x.

Think first, then show answer

log y = x log(x² + 1). Thus (1/y)dy/dx = log(x² + 1) + x × 2x/(x² + 1). Hence dy/dx = (x² + 1)^x[log(x² + 1) + 2x²/(x² + 1)].

Common wrong answer

Treating d/dx[x^x] as x×x^x−1, which incorrectly applies the power rule with a variable exponent.

Don't mix this up

Students take logarithms but forget to multiply the final derivative expression by y when converting from (1/y)dy/dx to dy/dx.

Exam tip

After differentiating log y, always write dy/dx = y times the right-hand side, then substitute the original value of y.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Common in long-answer differentiation problems involving x^x, powers with variable exponents, or lengthy products and quotients.

1
Enough for 1 mark
Logarithmic differentiation is a method where logarithms are taken on both sides before differentiating, especially for variable powers and complicated products or quotients.
2
Add for 2 marks
Example: If y = x^x, then log y = x log x. Differentiating gives (1/y)dy/dx = log x + 1, so dy/dx = x^x(log x + 1).
3
Add for 3 marks
Use this method when y has the form u(x)^v(x), or when y is a product and quotient of many factors.

MCQ Quiz

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