Logarithmic differentiation

What is logarithmic differentiation?

In plain words: the laws of logarithms turn powers into products, and products and quotients into sums and differences, which are much easier to differentiate than with the product and quotient rules. The left side becomes (1/y) dy/dx, by the chain rule. For y = u^v:

Here u and v are differentiable, u > 0, and primes denote their derivatives. In the implicit examples below, take positive variable bases and divide only where the coefficient of dy/dx is non-zero.

Products and quotients

For a long product or quotient, logarithms save repeated use of the product and quotient rules. NCERT Example 27 (page 131): y = √(P/Q), with P = (x − 3)(x² + 4) and Q = 3x² + 4x + 5. The real domain is x ≥ 3, and logarithmic differentiation here applies for x > 3. At x = 3 the square root has an unbounded right difference quotient. For x > 3,

log y = ½[log (x − 3) + log (x² + 4) − log (3x² + 4x + 5)],

so (1/y) dy/dx is ½ times the sum of three fractions: 1/(x − 3), then 2x/(x² + 4), then minus (6x + 4)/(3x² + 4x + 5). Multiply by y for dy/dx.

A variable in the exponent

Neither the power rule (for xⁿ, fixed n) nor the exponential rule (for a^x, fixed a) applies when both base and exponent vary. NCERT Example 29 (page 132): for y = x^{sin x}, x > 0, log y = sin x log x, so

(1/y) dy/dx = cos x log x + (sin x)/x,

and dy/dx = x^{sin x}[cos x log x + (sin x)/x].

The derivative of a^ˣ (NCERT Example 28, page 131): for a positive constant a, log y = x log a, so (1/y) dy/dx = log a:

With a = e this is e^ˣ again, since log e = 1.

Sums of such powers

Logarithms cannot be taken term by term: log (u + v) is not log u + log v. So for a sum, give each power a name and differentiate each separately. NCERT Example 30 (pages 132–133): for y^x + x^y + x^x = a^b, put u = y^x, v = x^y, w = x^x, so that u′ + v′ + w′ = 0, where

• log u = x log y gives u′ = y^x[log y + (x/y) dy/dx];
• log v = y log x gives v′ = x^y[y/x + log x dy/dx];
• log w = x log x gives w′ = x^x(1 + log x).

Collecting the dy/dx terms gives NCERT’s answer, dy/dx = −N/M, with N = y^x log y + yx^{y − 1} + x^x(1 + log x) and M = xy^{x − 1} + x^y log x.

Solved examples

Common mistakes to avoid

Quick revision summary

• For y = u^ᵛ, with u = u(x) > 0 and v = v(x), take logarithms: log y = v log u, then (1/y) dy/dx = v u′/u + v′ log u.
• Logarithms also turn long products and quotients into sums before differentiating.
• d/dx (a^ˣ) = a^ˣ log a and d/dx (x^ˣ) = x^ˣ(1 + log x); for sums of powers, differentiate each power separately.

Questions students ask

What is the derivative of x^xx?

The derivative of y = x^xx, for x > 0, is y × x^x times the sum of 1/x and (1 + log x) log x, where the tower means x raised to the power x^x, not (x^x)^x, which is the different function x^x2. Taking logarithms gives log y = x^x log x; the product rule, with (x^x)′ = x^x(1 + log x), then makes y′/y equal to x^x times that sum. At x = 2, where y = 16 and x^x = 4, the sum is about 1.674, so the derivative is 64 × 1.674 ≈ 107.1.

How do you differentiate the infinite power tower x^xx…?

For the infinite power tower y = x^xx…, the exponent of the first x is the whole tower again, so y = x^y, and, in the interior of its convergence interval, the derivative is y²/(x(1 − y log x)). Taking logarithms gives log y = y log x; differentiating implicitly, y′/y = y′ log x + y/x, and solving for y′ gives the result. Beyond NCERT, the tower is known to converge only for x from e^−e ≈ 0.066 to e^1/e ≈ 1.445; at x = √2 it settles at 2, where (√2)² = 2 (not at 4, the other solution of (√2)^y = y), and there the slope is 4/(√2(1 − log 2)), about 9.218. At the upper endpoint x = e^1/e, y = e makes the denominator zero, so there is no finite endpoint derivative; convergence alone does not guarantee differentiability.

How can you find the derivative of x^x without logarithmic differentiation?

You can find the derivative of x^x without logarithmic differentiation by writing x^x = e^{x log x}, true for x > 0 because e^{log x} = x, and using the chain rule: the derivative is e^{x log x} times (1 + log x), that is x^x(1 + log x), the same result. This is the alternative route NCERT shows for a^x in Example 28 (page 131), and it works for any u^v with u > 0, since u^v = e^{v log u}. As a check at x = 3, the slope of the chord from x = 2.999 to x = 3.001 is about 56.66, and 27(1 + log 3) ≈ 56.66.

What is the difference between logarithmic and implicit differentiation?

Logarithmic differentiation is a special use of implicit differentiation: you replace y = f(x) by log y = log f(x) and differentiate that equation implicitly, the left side giving (1/y) dy/dx by the chain rule. Implicit differentiation in general works on an equation linking x and y that is not solved for y, such as xy + sin y = 1, with no logarithms. Taking logarithms is the extra step that turns a variable power, a product or a quotient into a sum, and an equation such as y^x = x^y (NCERT Exercise 5.5, Q13) needs both steps.

When should you use logarithmic differentiation?

Use logarithmic differentiation when x is in both the base and the exponent, as in x^x or (sin x)^{log x}, since neither the power rule nor the a^x rule applies there, and when a function is a long product or quotient of powers, such as (x + 1)³(x + 2)⁵ divided by (x + 3)⁴, where logarithms turn it into a sum. For a short product such as x² sin x, or a power with a fixed exponent such as (x² + 1)⁷, the product rule or the chain rule is quicker.

Does logarithmic differentiation work if the function takes negative values?

Logarithmic differentiation still works for a product or quotient that takes negative values, at every point where it is not 0, if you take log |y| instead of log y: the derivative of log |u| is u′/u for every u ≠ 0, so the same sum of u′/u terms comes out; NCERT asks for positive functions (page 130) only so that log y is defined. For example, y = (x − 1)(x − 4) is −2 at x = 2, and log |y| is the sum of log |x − 1| and log |x − 4|, so there y′/y = 1 − ½ = ½ and y′ = −1, which matches y′ = 2x − 5. A variable power is different: in u^v the base must stay positive, since a power such as (−2)^1/2 is not a real number.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Differentiate cos x cos 2x cos 3x (NCERT Exercise 5.5, Q1, page 134).

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   log y = log cos x + log cos 2x + log cos 3x (where the cosines are positive), so y′/y = −tan x − 2 tan 2x − 3 tan 3x. The product-times-tangents form is valid wherever all three cosines are non-zero, also using absolute-value logs for negative factors. An answer valid for every real x, including zeros of the cosines, is −sin x cos 2x cos 3x − 2 cos x sin 2x cos 3x − 3 cos x cos 2x sin 3x.

2. Q2. Differentiate (log x)^{cos x}, for x > 1 (NCERT Exercise 5.5, Q3, page 134).

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   log y = cos x log (log x). The product rule gives y′/y = A − B, with A = cos x/(x log x) and B = sin x log (log x). So y′ = (log x)^{cos x}(A − B).

3. Q3. Find dy/dx if x^y + y^x = 1 (NCERT Exercise 5.5, Q12, page 134).

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   The derivative of x^y is x^y(y/x + log x y′), and of y^x is y^x(log y + (x/y) y′). Their sum is 0, so y′ = −(yx^{y − 1} + y^x log y)/(x^y log x + xy^{x − 1}).

4. Q4. Find dy/dx if xy = e^{x − y} (NCERT Exercise 5.5, Q15, page 134).

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   Since xy = e^{x − y} > 0, both variables are non-zero and have the same sign. Take logarithms: log |x| + log |y| = x − y. Then 1/x + y′/y = 1 − y′, so y′(1/y + 1) = 1 − 1/x, and y′ = y(x − 1)/(x(y + 1)), where y ≠ −1; this also covers the negative-variable branch.