MathsCBSE Class 12 · Continuity and Differentiability

Logarithmic Differentiation: Derivatives of x^x, u^v and Long Products (Class 12)

Updated 15 October 2026 · Fact-checked

Logarithmic differentiation means taking log on both sides before differentiating. For y = f(x)^g(x), write log y = g(x) log f(x), differentiate to get (1/y) dy/dx, then multiply by y. It is also the quickest way to differentiate long products and quotients. For a sum like x^x + (sin x)^x, differentiate each term separately.

When base and exponent both vary, take log first: the power becomes a product, then differentiate and multiply back by y.

Illustration drawn for QuestionBuddy.

Understand Logarithmic Differentiation

The power rule $\frac{d}{d x} x^{n} = n x^{n - 1}$ needs a constant exponent, and the rule $\frac{d}{d x} a^{x} = a^{x} lo g a$ needs a constant base. When both the base and the exponent vary, as in $x^{x}$ , $x^{s i n x}$ or $(lo g x)^{c o s x}$ , neither rule applies. Taking logarithms turns the power into a product, which we can differentiate.

**The method for $y = [f (x)]^{g (x)}$ ** (with $f (x) > 0$ ):

1. Take natural log: $lo g y = g (x) lo g f (x)$ .
2. Differentiate both sides with respect to $x$ ; the left side gives $\frac{1}{y} \frac{d y}{d x}$ by the chain rule, and the right side needs the product rule: $g^{'} (x) lo g f (x) + g (x) \frac{f ^{'} ( x )}{f ( x )}$ .
3. Multiply by $y$ and substitute $y$ back:
$\frac{d y}{d x} = [f (x)]^{g (x)} [g^{'} (x) lo g f (x) + \frac{g ( x ) f ^{'} ( x )}{f ( x )}]$ .

For $y = x^{x}$ : $lo g y = x lo g x$ , so $\frac{1}{y} y^{'} = lo g x + 1$ and $\frac{d y}{d x} = x^{x} (1 + lo g x)$ .

Long products and quotients. For something like $y = \frac{( x ^{2} + 1 ) ^{3} 2 x - 1}{( x + 2 ) ^{4}}$ , the quotient rule would be painful. Logs split it into a sum: $lo g y = 3 lo g (x^{2} + 1) + \frac{1}{2} lo g (2 x - 1) - 4 lo g (x + 2)$ , and each piece differentiates in one line.

Sums of such functions. $lo g$ does not split a sum: $lo g (u + v) \neq = lo g u + lo g v$ . For $y = u + v$ , find $\frac{d u}{d x}$ and $\frac{d v}{d x}$ separately (each by logarithmic differentiation if needed) and add. Setting up $u$ and $v$ clearly is worth a mark in the board exam.

Implicit equations with powers. Equations like $x^{y} \cdot y^{x} = 16$ or $y^{x} = e^{y - x}$ become, after taking logs, $y lo g x + x lo g y = lo g 16$ or $x lo g y = y - x$ , which you then differentiate implicitly.

Before you start, remember the derivatives you will use: $\frac{d}{d x} lo g x = \frac{1}{x}$ , $\frac{d}{d x} e^{x} = e^{x}$ , $\frac{d}{d x} a^{x} = a^{x} lo g a$ , and the chain rule. The logarithm here is the natural log (base $e$ ). This topic is NCERT Section 5.5 (with exponential and log derivatives in 5.4); it appears in most board papers, often combined with implicit or second-order derivatives.

Logarithmic Differentiation: formulas to remember

Variable base and exponent: $y = f^{g} \Rightarrow \frac{d y}{d x} = f^{g} (g^{'} lo g f + \frac{g f ^{'}}{f})$; Take log, differentiate, multiply by y.
x to the power x: $\frac{d}{d x} x^{x} = x^{x} (1 + lo g x)$; A standard result worth remembering.
Log of a product or quotient: $lo g \frac{uv}{w} = lo g u + lo g v - lo g w$; Turns long products into sums.
Log of a power: $lo g u^{n} = n lo g u$; Also for roots: $lo g u = \frac{1}{2} lo g u$ .
Basic derivatives: $\frac{d}{d x} lo g x = \frac{1}{x}, \frac{d}{d x} a^{x} = a^{x} lo g a, \frac{d}{d x} e^{x} = e^{x}$; Natural log throughout.
Sum of two such functions: $y = u + v \Rightarrow \frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$; Never take log of the whole sum.

How to solve Logarithmic Differentiation questions

Use this for any function with a variable exponent, any long product or quotient, and implicit equations with powers.

1If y is a sum of terms, write $y = u + v$ and handle each term separately.
2For each term of the form $[f (x)]^{g (x)}$ or each long product, take natural log of both sides and simplify with the log laws.
3Differentiate both sides with respect to $x$ : the left side becomes $\frac{1}{u} \frac{d u}{d x}$ ; use the product and chain rules on the right.
4Multiply by the original function to get $\frac{d u}{d x}$ , writing the function out in full.
5Add the parts: $\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$ .
6For implicit equations, take logs, differentiate implicitly, collect the $\frac{d y}{d x}$ terms and solve.
7If a value is asked, substitute at the end; keep exact forms such as $lo g 2$ .

Logarithmic Differentiation in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Derivative of a variable power

Suppose a bank's interest of 100% a year is credited in $n$ equal instalments, so Rs 1 grows to $f (n) = (1 + \frac{1}{n})^{n}$ rupees in a year. Treat $n$ as a continuous variable.

Find $f^{'} (n)$ by logarithmic differentiation and its value at $n = 1$ . Does crediting more often help?

Try it first, then show the solutionSolution

$lo g f = n lo g (1 + \frac{1}{n}) = n [lo g (n + 1) - lo g n]$ .
$\frac{f ^{'}}{f} = lo g (n + 1) - lo g n + n (\frac{1}{n + 1} - \frac{1}{n}) = lo g (1 + \frac{1}{n}) - \frac{1}{n + 1}$ .
At $n = 1$ : $f (1) = 2$ and $f^{'} (1) = 2 (lo g 2 - \frac{1}{2}) = 2 lo g 2 - 1 \approx 0.386$ .
The derivative is positive, so crediting more often increases the amount (towards $e \approx 2.718$ ).

Answer: $f^{'} (n) = (1 + \frac{1}{n})^{n} [lo g (1 + \frac{1}{n}) - \frac{1}{n + 1}]$ ; $f^{'} (1) = 2 lo g 2 - 1 \approx 0.386 > 0$ , so yes.

Memory hook: A power with n in the base and the exponent: take log first.

Relative rate of change of a product and quotient

Imagine a start-up models its weekly sales (in hundreds) as $S (x) = \frac{( x + 1 ) ^{2} ( x + 4 )}{x + 2}$ , where $x$ is the advertising budget in lakh rupees.

Use logarithmic differentiation to find the relative rate $\frac{S ^{'}}{S}$ and the rate $S^{'}$ when $x = 2$ .

Try it first, then show the solutionSolution

$lo g S = 2 lo g (x + 1) + lo g (x + 4) - lo g (x + 2)$ .
$\frac{S ^{'}}{S} = \frac{2}{x + 1} + \frac{1}{x + 4} - \frac{1}{x + 2}$ .
At $x = 2$ : $\frac{S ^{'}}{S} = \frac{2}{3} + \frac{1}{6} - \frac{1}{4} = \frac{8 + 2 - 3}{12} = \frac{7}{12}$ .
$S (2) = \frac{9 \times 6}{4} = 13.5$ , so $S^{'} (2) = 13.5 \times \frac{7}{12} = 7.875$ .

Answer: $\frac{S ^{'}}{S} = \frac{7}{12}$ at $x = 2$ , and $S^{'} (2) = 7.875$ hundred units per lakh rupees.

Memory hook: Logs turn a messy quotient into a sum of simple fractions.

Tricks to remember Logarithmic Differentiation

Short memory aids. Say each one aloud once, then use it on a question the same day.

Variable up and down? Log it
If $x$ is in both the base and the exponent, take log first. If only in the base, use the power rule; if only in the exponent, use $a^{x} lo g a$ .
Use it when: Choosing the method at a glance.
Log a sum? Never
$lo g (u + v)$ does not split. Break the sum into $u + v$ before logging.
Use it when: Questions like $x^{x} + (sin x)^{c o s x}$ .
Multiply back by y
The left side after differentiating is $\frac{1}{y} y^{'}$ , so the last step is always 'multiply by y and write y in full'.
Use it when: Not losing the final mark.
x^x(1 + log x)
Learn it as 'itself times one plus log': $x^{x} (1 + lo g x)$ .
Use it when: MCQs and as a building block.

Logarithmic Differentiation: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Variable base and exponent · 2 marks · Very short answerBoard level

Differentiate $y = x^{s i n x}$ , $x > 0$ , with respect to $x$ .

Try it first, then show the solutionSolution

$lo g y = sin x lo g x$ . (½ mark)
$\frac{1}{y} \frac{d y}{d x} = cos x lo g x + \frac{s i n x}{x}$ . (1 mark)
$\frac{d y}{d x} = x^{s i n x} (cos x lo g x + \frac{s i n x}{x})$ . (½ mark)

Answer: $\frac{d y}{d x} = x^{s i n x} (cos x lo g x + \frac{s i n x}{x})$

Example 2: A long product and quotient · 3 marks · Short answerBoard level

Find $\frac{d y}{d x}$ if $y = \frac{( x ^{2} + 1 ) ^{3} 2 x - 1}{( x + 2 ) ^{4}}$ , $x > \frac{1}{2}$ .

Try it first, then show the solutionSolution

$lo g y = 3 lo g (x^{2} + 1) + \frac{1}{2} lo g (2 x - 1) - 4 lo g (x + 2)$ . (1 mark)
$\frac{1}{y} \frac{d y}{d x} = \frac{6 x}{x ^{2} + 1} + \frac{1}{2 x - 1} - \frac{4}{x + 2}$ . (1½ marks)
$\frac{d y}{d x} = \frac{( x ^{2} + 1 ) ^{3} 2 x - 1}{( x + 2 ) ^{4}} [\frac{6 x}{x ^{2} + 1} + \frac{1}{2 x - 1} - \frac{4}{x + 2}]$ . (½ mark)

Answer: $\frac{d y}{d x} = y [\frac{6 x}{x ^{2} + 1} + \frac{1}{2 x - 1} - \frac{4}{x + 2}]$ with $y$ written in full.

Example 3: A sum of two variable powers · 3 marks · Short answerHarder board level

Differentiate $y = (1 + x)^{x} + x^{1 + x}$ , $x > 0$ .

Try it first, then show the solutionSolution

Let $u = (1 + x)^{x}$ and $v = x^{1 + x}$ , so $y = u + v$ . (½ mark)
$lo g u = x lo g (1 + x) \Rightarrow \frac{d u}{d x} = (1 + x)^{x} [lo g (1 + x) + \frac{x}{1 + x}]$ . (1 mark)
$lo g v = (1 + x) lo g x \Rightarrow \frac{d v}{d x} = x^{1 + x} [lo g x + \frac{1 + x}{x}]$ . (1 mark)
$\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$ . (½ mark)

Answer: $\frac{d y}{d x} = (1 + x)^{x} [lo g (1 + x) + \frac{x}{1 + x}] + x^{1 + x} [lo g x + \frac{1 + x}{x}]$

Example 4: Implicit equation with powers · 3 marks · Short answerHarder board level

If $x^{y} \cdot y^{x} = 16$ , find $\frac{d y}{d x}$ and show that it equals $- 1$ at $(2, 2)$ .

Try it first, then show the solutionSolution

Take logs: $y lo g x + x lo g y = lo g 16$ . (½ mark)
Differentiate: $\frac{d y}{d x} lo g x + \frac{y}{x} + lo g y + \frac{x}{y} \frac{d y}{d x} = 0$ . (1 mark)
$\frac{d y}{d x} (lo g x + \frac{x}{y}) = - (lo g y + \frac{y}{x})$ , so $\frac{d y}{d x} = - \frac{l o g y + \frac{y}{x}}{l o g x + \frac{x}{y}}$ . (1 mark)
$(2, 2)$ is on the curve ( $4 \times 4 = 16$ ); there $\frac{d y}{d x} = - \frac{l o g 2 + 1}{l o g 2 + 1} = - 1$ . (½ mark)

Answer: $\frac{d y}{d x} = - \frac{l o g y + \frac{y}{x}}{l o g x + \frac{x}{y}}$ ; at $(2, 2)$ it is $- 1$ .

Example 5: Two tough terms · 5 marks · Long answerHOTS / competency

Differentiate $y = (x sin x)^{x} + (sin x)^{x}$ , $0 < x < π$ .

Try it first, then show the solutionSolution

Let $u = (x sin x)^{x}$ , $v = (sin x)^{x}$ . (½ mark)
$lo g u = x [lo g x + lo g sin x]$ . Differentiating: $\frac{1}{u} \frac{d u}{d x} = lo g x + lo g sin x + x (\frac{1}{x} + cot x)$ . (1½ marks)
So $\frac{d u}{d x} = (x sin x)^{x} [lo g (x sin x) + 1 + x cot x]$ . (½ mark)
$lo g v = x lo g sin x$ , so $\frac{1}{v} \frac{d v}{d x} = \frac{l o g s i n x}{2 x} + x cot x$ . (1½ marks)
$\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$ . (1 mark)

Answer: $\frac{d y}{d x} = (x sin x)^{x} [lo g (x sin x) + 1 + x cot x] + (sin x)^{x} [\frac{l o g s i n x}{2 x} + x cot x]$

Example 6: Case study: a fast-growing culture · 4 marks · Case-basedHOTS / competency

Suppose a lab models the size of a bacterial culture (in thousands) by $N (t) = (t + 1)^{t}$ , where $t \geq 0$ is in hours.
(i) Find $N^{'} (t)$ using logarithmic differentiation. (2 marks)
(ii) Find the growth rate at $t = 1$ and $t = 2$ . (1 mark)
(iii) Show that the relative growth rate $\frac{N ^{'}}{N}$ increases with $t$ . (1 mark)

Try it first, then show the solutionSolution

(i) $lo g N = t lo g (t + 1)$ , so $\frac{N ^{'}}{N} = lo g (t + 1) + \frac{t}{t + 1}$ and $N^{'} (t) = (t + 1)^{t} [lo g (t + 1) + \frac{t}{t + 1}]$ . (2 marks)
(ii) $N^{'} (1) = 2 (lo g 2 + \frac{1}{2}) = 2 lo g 2 + 1 \approx 2.39$ thousand per hour; $N^{'} (2) = 9 (lo g 3 + \frac{2}{3}) = 9 lo g 3 + 6 \approx 15.89$ thousand per hour. (1 mark)
(iii) Both $lo g (t + 1)$ and $\frac{t}{t + 1} = 1 - \frac{1}{t + 1}$ increase as $t$ increases, so their sum $\frac{N ^{'}}{N}$ increases. (1 mark)

Answer: (i) $N^{'} = (t + 1)^{t} [lo g (t + 1) + \frac{t}{t + 1}]$ (ii) about 2.39 and 15.89 thousand per hour (iii) both parts increase.

Where students lose marks in Logarithmic Differentiation

The slips examiners see most often, and what to write instead.

Mistake: Using the power rule on x^x
The exponent is not constant.
Wrong working
$\frac{d}{d x} x^{s i n x} = sin x \cdot x^{s i n x - 1}$ .
Right working
$\frac{d}{d x} x^{s i n x} = x^{s i n x} (cos x lo g x + \frac{s i n x}{x})$ .
Do this: Check: is the exponent a function of x? Then take logs.
Usually costs: The whole question.
Mistake: Taking log of a sum
$lo g (u + v) \neq = lo g u + lo g v$ .
Wrong working
$y = x^{x} + 2^{x} \Rightarrow lo g y = x lo g x + x lo g 2$ .
Right working
Differentiate $u = x^{x}$ and $v = 2^{x}$ separately: $y^{'} = x^{x} (1 + lo g x) + 2^{x} lo g 2$ .
Do this: Split sums first, then log each power term.
Usually costs: All marks.
Mistake: Forgetting to multiply by y
Differentiating $lo g y$ gives $\frac{1}{y} y^{'}$ , not $y^{'}$ .
Wrong working
$lo g y = x lo g x \Rightarrow \frac{d y}{d x} = 1 + lo g x$ .
Right working
$\frac{1}{y} \frac{d y}{d x} = 1 + lo g x \Rightarrow \frac{d y}{d x} = x^{x} (1 + lo g x)$ .
Do this: End every log-differentiation with 'multiply by y and substitute'.
Usually costs: The final mark.
Mistake: Missing the product rule on the right side
$g (x) lo g f (x)$ is a product of two functions.
Wrong working
$lo g y = sin x lo g x \Rightarrow \frac{y ^{'}}{y} = cos x \cdot \frac{1}{x}$ .
Right working
$\frac{y ^{'}}{y} = cos x lo g x + sin x \cdot \frac{1}{x}$ .
Do this: Write the product rule explicitly for the right side.
Usually costs: 1 to 2 marks.

Exam tips

Logarithmic differentiation appeared in the 2023, 2024 and 2026 board papers (including a 5-mark question in 2026); show the line with $lo g y$ clearly.
For sums, define $u$ and $v$ at the start; examiners award a mark for $\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$ .
Write the final answer with the original function in full, not just $y$ , unless the question allows it.
Use log laws to simplify before differentiating; it shortens the work and reduces errors.
State any condition such as $x > 0$ or $sin x > 0$ needed for the logarithm to be defined.

Practice questions on Logarithmic Differentiation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$\frac{d}{d x} (x^{x})$ equals:
(a) $x \cdot x^{x - 1}$
(b) $x^{x} lo g x$
(c) $x^{x} (1 + lo g x)$
(d) $x^{x}$
Hint
Take log: $lo g y = x lo g x$ .
Answer
(c) $x^{x} (1 + lo g x)$ .
Q2 · 2 marks · Very short answerBoard level
Differentiate $y = (cos x)^{l o g x}$ , $0 < x < \frac{π}{2}$ .
Hint
$lo g y = lo g x \cdot lo g cos x$ .
Answer
$\frac{1}{y} y^{'} = \frac{l o g c o s x}{x} - lo g x tan x$ , so $\frac{d y}{d x} = (cos x)^{l o g x} [\frac{l o g c o s x}{x} - tan x lo g x]$ .
Q3 · 2 marks · Very short answerBoard level
Differentiate $y = x^{1/ x}$ , $x > 0$ .
Hint
$lo g y = \frac{l o g x}{x}$ ; use the quotient rule.
Answer
$\frac{1}{y} y^{'} = \frac{1 - l o g x}{x ^{2}}$ , so $\frac{d y}{d x} = x^{1/ x} \cdot \frac{1 - l o g x}{x ^{2}}$ .
Q4 · 3 marks · Short answerHarder board level
Differentiate $y = \frac{( x - 1 ) ( x + 2 )}{( x + 3 ) ( x - 4 )}$ , $x > 4$ .
Hint
The square root becomes a factor $\frac{1}{2}$ after taking logs.
Answer
$lo g y = \frac{1}{2} [lo g (x - 1) + lo g (x + 2) - lo g (x + 3) - lo g (x - 4)]$ , so $\frac{d y}{d x} = \frac{y}{2} [\frac{1}{x - 1} + \frac{1}{x + 2} - \frac{1}{x + 3} - \frac{1}{x - 4}]$ with $y$ written in full.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\frac{d}{d x} (x^{x}) = x \cdot x^{x - 1}$ .
Reason (R): $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ holds when n is a constant.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Is the exponent in $x^{x}$ a constant?
Answer
(d). The power rule needs a constant exponent, so it cannot be applied to $x^{x}$ ; the correct derivative is $x^{x} (1 + lo g x)$ . A is false, R is true.
Q6 · 3 marks · Short answerHarder board level
If $y^{x} = e^{y - x}$ , prove that $\frac{d y}{d x} = \frac{y ( 1 + l o g y )}{y - x}$ .
Hint
Take logs: $x lo g y = y - x$ .
Answer
$x lo g y = y - x$ . Differentiating: $lo g y + \frac{x}{y} y^{'} = y^{'} - 1$ , so $y^{'} (1 - \frac{x}{y}) = 1 + lo g y$ , i.e. $y^{'} \cdot \frac{y - x}{y} = 1 + lo g y$ and $\frac{d y}{d x} = \frac{y ( 1 + l o g y )}{y - x}$ .
Q7 · 3 marks · Short answerHOTS / competency
For $y = x^{x}$ , $x > 0$ : (i) find $\frac{d y}{d x}$ ; (ii) find where $\frac{d y}{d x} = 0$ ; (iii) find the value of $y$ there, correct to three decimal places.
Hint
$1 + lo g x = 0$ gives $x = e^{- 1}$ .
Answer
(i) $x^{x} (1 + lo g x)$ . (ii) $x^{x} > 0$ , so $\frac{d y}{d x} = 0$ when $lo g x = - 1$ , i.e. $x = \frac{1}{e}$ . (iii) $y = (\frac{1}{e})^{1/ e} = e^{- 1/ e} \approx 0.692$ (this is the least value of $x^{x}$ ).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is $\frac{d}{d x} x^{x}$ ?Show answer
$x^{x} (1 + lo g x)$ .
2.What is the first step for $y = (sin x)^{c o s x}$ ?Show answer
Take logs: $lo g y = cos x lo g sin x$ .
3.Why can't you take log of $x^{x} + x^{2}$ directly?Show answer
Log does not split sums; differentiate the terms separately.
4.What does $\frac{d}{d x} lo g y$ equal?Show answer
$\frac{1}{y} \frac{d y}{d x}$ .
5.What is $\frac{d}{d x} 2^{x}$ ?Show answer
$2^{x} lo g 2$ .
6.Differentiate $x^{2 x}$ .Show answer
$x^{2 x} (2 lo g x + 2)$ .

Logarithmic Differentiation: frequently asked questions

When should I use logarithmic differentiation?

When the variable appears in both the base and the exponent (like x^x or (sin x)^x), or when a function is a long product or quotient of powers. Taking logs turns powers into products and products into sums.

How do I differentiate x^x + (sin x)^x?

Split it: u = x^x and v = (sin x)^x. Differentiate each using logs, then add: dy/dx = du/dx + dv/dx. Never take the log of the whole sum.

Is log here the natural logarithm?

Yes. In Class 12 calculus, log x means the natural logarithm (base e), so its derivative is 1/x.

Which NCERT section covers logarithmic differentiation?

Section 5.5 (Logarithmic Differentiation) of Chapter 5, Continuity and Differentiability, in NCERT Mathematics Part I for Class 12.

What is the derivative of a^x?

a^x log a for a constant a > 0. Only when both base and exponent vary do you need logarithmic differentiation.