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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.

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Student-friendly explanation

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.

How to write this in exams

  1. 1

    Start with the exact idea

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

  2. 2

    Then show how to use it

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

  3. 3

    Add one concrete 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).

  4. 4

    Avoid this incomplete answer

    Treating d/dx[x^x] as x*x^(x-1), which incorrectly applies the power rule with a variable exponent.

Definition

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

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).

Rule to remember

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.

Memory hook

Log first, expand neatly, differentiate, multiply back by y.

Examples and method

Worked example

Differentiate y = x^2 sqrt(x + 1)/(x - 3), where factors are positive in the working interval. Taking logs: log y = 2log x + (1/2)log(x + 1) - log(x - 3). Differentiate: (1/y)dy/dx = 2/x + 1/[2(x + 1)] - 1/(x - 3). Therefore dy/dx = [x^2 sqrt(x + 1)/(x - 3)][2/x + 1/(2x + 2) - 1/(x - 3)].

Method to apply

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

Diagram support

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

How CBSE asks it

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

Avoid common mistakes

Common confusion

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

Common wrong answer

Treating d/dx[x^x] as x*x^(x-1), which incorrectly applies the power rule with a variable exponent.

Exam tip

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

Quick check

Differentiate y = (x^2 + 1)^x.

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

Answer writing and exam use

1-mark answer

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

2-mark answer

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

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. 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. Differentiate y = x^2 sqrt(x + 1)/(x - 3), where factors are positive in the working interval. Taking logs: log y = 2log x + (1/2)log(x + 1) - log(x - 3). Differentiate: (1/y)dy/dx = 2/x + 1/[2(x + 1)] - 1/(x - 3). Therefore dy/dx = [x^2 sqrt(x + 1)/(x - 3)][2/x + 1/(2x + 2) - 1/(x - 3)]. Common in long-answer differentiation problems involving x^x, powers with variable exponents, or lengthy products and quotients. Treating d/dx[x^x] as x*x^(x-1), which incorrectly applies the power rule with a variable exponent.
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