When should I use logarithmic differentiation?

When x is in both the base and the exponent, or when a product or quotient has three or more factors with powers or roots.

Is logarithmic differentiation the same as implicit differentiation?

It is a special case. Taking logs creates the equation ln y = ln f(x), and you then differentiate it implicitly, with d/dx ln y = y'/y.

Why does logarithmic differentiation work?

Because ln turns the hard operations into easy ones: products into sums, quotients into differences, powers into multiples. Differentiating a sum is far easier than differentiating a product, and the chain rule on ln y brings back the y you need at the end.

What is the derivative of x^x?

x^x(ln x + 1), for x > 0. It is the first example above.

Do I need the quotient rule if I use logarithmic differentiation?

No, which is half the point. In our answer data the quotient rule is right 76% of the time against 84% for the product rule; logs turn quotients into subtraction and sidestep it.

Where can I find logarithmic differentiation practice problems?

Right on this page: the practice problems section has three logarithmic differentiation problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 500 related problems after that, free and with no account: try a five-minute Sprint.

Differentiate, for x > 0, without the quotient rule: y = ((x^2+1)^3)/(√x)

((x^2+1)^3)/(√x)·(6x/(x^2+1)-1/2x)

Blog · Calculus guides · 5 November 2026 · 4 min read

By the Asymptote team. Every answer on this page is checked by the same machine that grades the game.

Logarithmic Differentiation: Steps, x^x & Practice Problems

An illustration of Logarithmic differentiation

Logarithmic differentiation means taking $ln$ of both sides of $y = f (x)$ , using the log laws to break products, quotients and powers apart, then differentiating implicitly. It is the standard way to differentiate a variable raised to a variable power, like $x^{x}$ , and a shortcut for long products and quotients. Since $\frac{d}{d x} ln y = \frac{y ^{'}}{y}$ , you finish by multiplying through by $y$ .

The formula

ln y = ln f (x) ⟹ \frac{y ^{'}}{y} = \frac{d}{d x} ln f (x) ⟹ y^{'} = f (x) \cdot \frac{d}{d x} ln f (x)

Logs turn multiplication into addition and powers into multiplication. You spend one line taking logs and save five lines of quotient rule.

When to use it

Use it in two situations. First, whenever $x$ appears in both the base and the exponent: $x^{x}$ , $x^{s i n x}$ , $(ln x)^{x}$ . Neither the power rule (which needs a constant exponent) nor the rule for $a^{x}$ (which needs a constant base) applies, and logs are the way in. Second, for a product or quotient of three or more factors with powers and roots, where the quotient rule would be long and error-prone. The method rests on implicit differentiation: the left side becomes $\frac{y ^{'}}{y}$ by the chain rule.

How to do it, step by step

Take $ln$ of both sides. $y = f (x)$ becomes $ln y = ln f (x)$ . You need $y > 0$ for this; for the derivative it is fine to use $ln ∣ y ∣$ and get the same answer.
Expand with the log laws. $ln (ab) = ln a + ln b$ , $ln \frac{a}{b} = ln a - ln b$ , and $ln a^{k} = k ln a$ . Pull every power down to the front.
Differentiate both sides. The left side gives $\frac{y ^{'}}{y}$ . The right side is now a sum of simple logs, each differentiated with the chain rule.
Multiply by $y$ and substitute it back. Replace $y$ with the original function so the answer is in $x$ alone.

Worked examples

Example 1: The classic: x to the x

Easy

Differentiate, for $x > 0$ :

y = x^{x}

Take logs and bring the exponent down: $ln y = ln (x^{x}) = x ln x$ .

Differentiate both sides. The right side is a product, $x$ times $ln x$ :

\frac{y ^{'}}{y} = 1 \cdot ln x + x \cdot \frac{1}{x} = ln x + 1

Multiply by $y = x^{x}$ to get $y^{'}$ on its own. The tempting wrong answer, $x \cdot x^{x - 1}$ , treats the exponent as a constant; it is not, which is the entire reason for this method.

Answer

\frac{d y}{d x} = x^{x} (ln x + 1)

Example 2: A trig function in the exponent

Medium

Differentiate, for $x > 0$ :

y = x^{s i n x}

Take logs: $ln y = sin x \cdot ln x$ .

Differentiate the right side with the product rule:

\frac{y ^{'}}{y} = cos x \cdot ln x + sin x \cdot \frac{1}{x}

Multiply both sides by $y = x^{s i n x}$ and you are done. Notice the shape: the original function, times the derivative of its logarithm. Every logarithmic differentiation answer looks like this.

You can also get here without logs by writing $x^{s i n x} = e^{s i n x l n x}$ and using the chain rule. It is the same calculation in a different coat, and some people find it easier to trust.

Answer

\frac{d y}{d x} = x^{s i n x} (cos x ln x + \frac{sin x}{x})

Example 3: A quotient without the quotient rule

Harder

Differentiate, for $x > 0$ :

y = \frac{x ^{2} 3 x + 1}{( x + 1 ) ^{3}}

Three factors, a root and two powers: by the quotient rule this is a page of algebra. Take logs instead and let the log laws split it up:

ln y = 2 ln x + \frac{1}{2} ln (3 x + 1) - 3 ln (x + 1)

Each term is now a one-line chain rule. The middle one gives $\frac{1}{2} \cdot \frac{3}{3 x + 1}$ .

\frac{y ^{'}}{y} = \frac{2}{x} + \frac{3}{2 ( 3 x + 1 )} - \frac{3}{x + 1}

Multiply by $y$ , written out in full. You can combine the bracket into one fraction if a question insists, but this form is correct and much harder to get wrong.

For comparison, the quotient rule here would need the product rule inside it for the top, plus a chain rule for the root, before any simplifying. Logs replaced all of that with three short derivatives and some addition.

Answer

\frac{d y}{d x} = \frac{x ^{2} 3 x + 1}{( x + 1 ) ^{3}} \cdot (\frac{2}{x} + \frac{3}{2 ( 3 x + 1 )} - \frac{3}{x + 1})

Logarithmic differentiation practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

Differentiate, for $x > 0$ :

y = x^{2 x}

Stuck? Show the answer

Answer

x^{2 x} (2 ln x + 2)

How to start

Take logs: $ln y = 2 x ln x$ . Differentiate the right side with the product rule, then multiply by $y$ .

Problem 2

Medium

Differentiate, for $x > 0$ :

y = x^{x}

Stuck? Show the answer

Answer

x^{x} \cdot \frac{ln x + 2}{2 x}

How to start

$ln y = x ln x$ . The product rule gives $\frac{l n x}{2 x} + \frac{x}{x}$ ; put both over $2 x$ before multiplying by $y$ .

Problem 3

Harder

Differentiate, for $x > 0$ , without the quotient rule:

y = \frac{( x ^{2} + 1 ) ^{3}}{x}

Stuck? Show the answer

Answer

\frac{( x ^{2} + 1 ) ^{3}}{x} \cdot (\frac{6 x}{x ^{2} + 1} - \frac{1}{2 x})

How to start

Logs split it: $ln y = 3 ln (x^{2} + 1) - \frac{1}{2} ln x$ . Differentiate each term with the chain rule, then multiply by $y$ .

Want more? Asymptote has 500 related problems, free, with no account.

Common mistakes

Using the power rule on $x^{x}$ . $x \cdot x^{x - 1}$ is wrong: the power rule needs a constant exponent. So is $x^{x} ln x$ , which treats the base as constant. The real answer contains both pieces: $x^{x} (ln x + 1)$ .
Forgetting to multiply by $y$ . After differentiating you have $\frac{y ^{'}}{y}$ , not $y^{'}$ . Stopping there gives the derivative of $ln y$ instead of the derivative of $y$ .
Breaking the log laws. $ln (a + b)$ is not $ln a + ln b$ . The laws split products and quotients, never sums, so $ln (3 x + 1)$ stays whole.
Leaving $y$ in the answer. Substitute the original function back in, so the derivative is in terms of $x$ only.
Confusing $x^{x}$ with $e^{x}$ . $\frac{d}{d x} e^{x} = e^{x}$ because the base is the constant $e$ . For $x^{x}$ the base moves too, so the derivative picks up the extra $ln x$ term.

Practice it now

Five minutes, no account

There are no logarithmic differentiation problems in Asymptote as such, so no $x^{x}$ yet. The closest thing is the derivative of $ln$ of a polynomial, the same chain-rule step this method uses on every term, which a Sprint serves from about the sixth problem.

Play a Sprint See it on the formula sheet

FAQ

When should I use logarithmic differentiation?: When $x$ is in both the base and the exponent, or when a product or quotient has three or more factors with powers or roots.
Is logarithmic differentiation the same as implicit differentiation?: It is a special case. Taking logs creates the equation $ln y = ln f (x)$ , and you then differentiate it implicitly, with $\frac{d}{d x} ln y = \frac{y ^{'}}{y}$ .
Why does logarithmic differentiation work?: Because $ln$ turns the hard operations into easy ones: products into sums, quotients into differences, powers into multiples. Differentiating a sum is far easier than differentiating a product, and the chain rule on $ln y$ brings back the $y$ you need at the end.
What is the derivative of $x^{x}$ ?: $x^{x} (ln x + 1)$ , for $x > 0$ . It is the first example above.
Do I need the quotient rule if I use logarithmic differentiation?: No, which is half the point. In our answer data the quotient rule is right 76% of the time against 84% for the product rule; logs turn quotients into subtraction and sidestep it.
Where can I find logarithmic differentiation practice problems?: Right on this page: the practice problems section has three logarithmic differentiation problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 500 related problems after that, free and with no account: try a five-minute Sprint.

Blog · Calculus guides · 5 November 2026 · 4 min read

By the Asymptote team. Every answer on this page is checked by the same machine that grades the game.

Logarithmic Differentiation: Steps, x^x & Practice Problems

The formula

ln y = ln f (x) ⟹ \frac{y ^{'}}{y} = \frac{d}{d x} ln f (x) ⟹ y^{'} = f (x) \cdot \frac{d}{d x} ln f (x)

Logs turn multiplication into addition and powers into multiplication. You spend one line taking logs and save five lines of quotient rule.

When to use it

How to do it, step by step

Take $ln$ of both sides. $y = f (x)$ becomes $ln y = ln f (x)$ . You need $y > 0$ for this; for the derivative it is fine to use $ln ∣ y ∣$ and get the same answer.
Expand with the log laws. $ln (ab) = ln a + ln b$ , $ln \frac{a}{b} = ln a - ln b$ , and $ln a^{k} = k ln a$ . Pull every power down to the front.
Differentiate both sides. The left side gives $\frac{y ^{'}}{y}$ . The right side is now a sum of simple logs, each differentiated with the chain rule.
Multiply by $y$ and substitute it back. Replace $y$ with the original function so the answer is in $x$ alone.

Worked examples

Example 1: The classic: x to the x

Easy

Differentiate, for $x > 0$ :

y = x^{x}

Take logs and bring the exponent down: $ln y = ln (x^{x}) = x ln x$ .

Differentiate both sides. The right side is a product, $x$ times $ln x$ :

\frac{y ^{'}}{y} = 1 \cdot ln x + x \cdot \frac{1}{x} = ln x + 1

Multiply by $y = x^{x}$ to get $y^{'}$ on its own. The tempting wrong answer, $x \cdot x^{x - 1}$ , treats the exponent as a constant; it is not, which is the entire reason for this method.

Answer

\frac{d y}{d x} = x^{x} (ln x + 1)

Example 2: A trig function in the exponent

Medium

Differentiate, for $x > 0$ :

y = x^{s i n x}

Take logs: $ln y = sin x \cdot ln x$ .

Differentiate the right side with the product rule:

\frac{y ^{'}}{y} = cos x \cdot ln x + sin x \cdot \frac{1}{x}

Multiply both sides by $y = x^{s i n x}$ and you are done. Notice the shape: the original function, times the derivative of its logarithm. Every logarithmic differentiation answer looks like this.

You can also get here without logs by writing $x^{s i n x} = e^{s i n x l n x}$ and using the chain rule. It is the same calculation in a different coat, and some people find it easier to trust.

Answer

\frac{d y}{d x} = x^{s i n x} (cos x ln x + \frac{sin x}{x})

Example 3: A quotient without the quotient rule

Harder

Differentiate, for $x > 0$ :

y = \frac{x ^{2} 3 x + 1}{( x + 1 ) ^{3}}

Three factors, a root and two powers: by the quotient rule this is a page of algebra. Take logs instead and let the log laws split it up:

ln y = 2 ln x + \frac{1}{2} ln (3 x + 1) - 3 ln (x + 1)

Each term is now a one-line chain rule. The middle one gives $\frac{1}{2} \cdot \frac{3}{3 x + 1}$ .

\frac{y ^{'}}{y} = \frac{2}{x} + \frac{3}{2 ( 3 x + 1 )} - \frac{3}{x + 1}

Multiply by $y$ , written out in full. You can combine the bracket into one fraction if a question insists, but this form is correct and much harder to get wrong.

Answer

\frac{d y}{d x} = \frac{x ^{2} 3 x + 1}{( x + 1 ) ^{3}} \cdot (\frac{2}{x} + \frac{3}{2 ( 3 x + 1 )} - \frac{3}{x + 1})

Logarithmic differentiation practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

Differentiate, for $x > 0$ :

y = x^{2 x}

Stuck? Show the answer

Answer

x^{2 x} (2 ln x + 2)

How to start

Take logs: $ln y = 2 x ln x$ . Differentiate the right side with the product rule, then multiply by $y$ .

Problem 2

Medium

Differentiate, for $x > 0$ :

y = x^{x}

Stuck? Show the answer

Answer

x^{x} \cdot \frac{ln x + 2}{2 x}

How to start

$ln y = x ln x$ . The product rule gives $\frac{l n x}{2 x} + \frac{x}{x}$ ; put both over $2 x$ before multiplying by $y$ .

Problem 3

Harder

Differentiate, for $x > 0$ , without the quotient rule:

y = \frac{( x ^{2} + 1 ) ^{3}}{x}

Stuck? Show the answer

Answer

\frac{( x ^{2} + 1 ) ^{3}}{x} \cdot (\frac{6 x}{x ^{2} + 1} - \frac{1}{2 x})

How to start

Logs split it: $ln y = 3 ln (x^{2} + 1) - \frac{1}{2} ln x$ . Differentiate each term with the chain rule, then multiply by $y$ .

Want more? Asymptote has 500 related problems, free, with no account.

Common mistakes

Using the power rule on $x^{x}$ . $x \cdot x^{x - 1}$ is wrong: the power rule needs a constant exponent. So is $x^{x} ln x$ , which treats the base as constant. The real answer contains both pieces: $x^{x} (ln x + 1)$ .
Forgetting to multiply by $y$ . After differentiating you have $\frac{y ^{'}}{y}$ , not $y^{'}$ . Stopping there gives the derivative of $ln y$ instead of the derivative of $y$ .
Breaking the log laws. $ln (a + b)$ is not $ln a + ln b$ . The laws split products and quotients, never sums, so $ln (3 x + 1)$ stays whole.
Leaving $y$ in the answer. Substitute the original function back in, so the derivative is in terms of $x$ only.
Confusing $x^{x}$ with $e^{x}$ . $\frac{d}{d x} e^{x} = e^{x}$ because the base is the constant $e$ . For $x^{x}$ the base moves too, so the derivative picks up the extra $ln x$ term.

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

When should I use logarithmic differentiation?: When $x$ is in both the base and the exponent, or when a product or quotient has three or more factors with powers or roots.
Is logarithmic differentiation the same as implicit differentiation?: It is a special case. Taking logs creates the equation $ln y = ln f (x)$ , and you then differentiate it implicitly, with $\frac{d}{d x} ln y = \frac{y ^{'}}{y}$ .
Why does logarithmic differentiation work?: Because $ln$ turns the hard operations into easy ones: products into sums, quotients into differences, powers into multiples. Differentiating a sum is far easier than differentiating a product, and the chain rule on $ln y$ brings back the $y$ you need at the end.
What is the derivative of $x^{x}$ ?: $x^{x} (ln x + 1)$ , for $x > 0$ . It is the first example above.
Do I need the quotient rule if I use logarithmic differentiation?: No, which is half the point. In our answer data the quotient rule is right 76% of the time against 84% for the product rule; logs turn quotients into subtraction and sidestep it.
Where can I find logarithmic differentiation practice problems?: Right on this page: the practice problems section has three logarithmic differentiation problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 500 related problems after that, free and with no account: try a five-minute Sprint.

Logarithmic Differentiation: Steps, x^x & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: The classic: x to the x

Example 2: A trig function in the exponent

Example 3: A quotient without the quotient rule

Logarithmic differentiation practice problems

Common mistakes

FAQ

Keep going

Logarithmic Differentiation: Steps, x^x & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: The classic: x to the x

Example 2: A trig function in the exponent

Example 3: A quotient without the quotient rule

Logarithmic differentiation practice problems

Common mistakes

FAQ

Keep going