Blog · Calculus guides · 24 October 2026 · 4 min read

Derivatives of e^x and ln x: Examples & Practice Problems

Q: Why is the derivative of e^x equal to e^x?

That is what makes e special. Every exponential a^x has a derivative proportional to itself, a^x ln a, and e ≈ 2.718 is the one base where that constant, ln e, is exactly 1.

Q: What is the derivative of ln(kx)?

Just 1/x, for any positive constant k. The chain rule gives k/kx, and the k cancels. Another way to see it: ln(kx) = ln k + ln x, and ln k is a constant.

Q: What are the integrals of e^x and 1/x?

Read the derivatives backwards: ∫ e^x dx = e^x + C and ∫1/x dx = ln|x| + C. The absolute value is there because ln|x| also has derivative 1/x for negative x, where plain ln x does not exist. With an inside, divide by its derivative instead of multiplying, as long as it is a constant: ∫ e^3x dx = 1/3e^3x + C.

Q: What about other bases, like 2^x and log base 2?

d/dxa^x = a^x ln a, and d/dx log_a x = 1/(x ln a). With a = e the ln a is 1 and you get back the rules above.

An illustration of Derivatives of e^x and ln x

The derivative of $e^{x}$ is $e^{x}$ , and the derivative of $ln x$ is $\frac{1}{x}$ (for $x > 0$ ). When there is something other than plain $x$ inside, multiply by its derivative: $\frac{d}{d x} e^{g (x)} = g^{'} (x) e^{g (x)}$ and $\frac{d}{d x} ln g (x) = \frac{g ^{'} ( x )}{g ( x )}$ . For example, $\frac{d}{d x} e^{3 x} = 3 e^{3 x}$ and $\frac{d}{d x} ln (x^{2} + 1) = \frac{2 x}{x ^{2} + 1}$ .

The formula

\frac{d}{d x} [e^{x}] = e^{x} \frac{d}{d x} [ln x] = \frac{1}{x} \frac{d}{d x} [e^{g}] = g^{'} e^{g} \frac{d}{d x} [ln g] = \frac{g ^{'}}{g}

$e^{x}$ is the function that is its own derivative, which is the whole reason $e$ is the number it is. Differentiate it all afternoon and it will not change, which is either peaceful or unsettling depending on the afternoon.

When to use it

Use these whenever $e$ is raised to a power with $x$ in it, or you are taking the natural log of something. The plain versions, $e^{x}$ and $ln x$ , are straight off the table. Anything else, $e^{3 x}$ , $e^{- x^{2}}$ , $ln (5 x)$ , $ln (x^{2} + 1)$ , needs the chain rule on top: the derivative of the inside comes out as a factor. A product like $x^{2} e^{x}$ needs the product rule as well. And for a log of a big product or quotient, it is often easiest to split it with the log laws before you differentiate anything.

How to do it, step by step

Find the inside. In $e^{3 x^{2}}$ the inside is $3 x^{2}$ ; in $ln (x^{2} + 1)$ it is $x^{2} + 1$ . If the inside is just $x$ , the table rule is the whole answer.
For $e^{□}$ : copy it, then multiply by the inside's derivative. $e^{3 x^{2}}$ stays $e^{3 x^{2}}$ and gains a factor of $6 x$ .
For $ln (□)$ : put the inside's derivative over the inside. $ln (x^{2} + 1)$ becomes $\frac{2 x}{x ^{2} + 1}$ .
Use the log laws first if they help. $ln (x^{3}) = 3 ln x$ and $ln \frac{a}{b} = ln a - ln b$ often turn a hard derivative into an easy one.

Worked examples

Example 1: Straight from the table

Easy

Differentiate:

y = 5 e^{x} - 3 ln x

Constants stay in front. $e^{x}$ differentiates to itself, and $ln x$ to $\frac{1}{x}$ .

\frac{d y}{d x} = 5 \cdot e^{x} - 3 \cdot \frac{1}{x}

Answer

\frac{d y}{d x} = 5 e^{x} - \frac{3}{x}

Example 2: With something inside

Medium

Differentiate:

y = 4 e^{3 x} + ln (x^{2} + 1)

First term: the inside is $3 x$ , whose derivative is 3. So $4 e^{3 x}$ becomes $4 \cdot 3 e^{3 x}$ . The exponent itself does not change.

Second term: the inside is $x^{2} + 1$ , whose derivative is $2 x$ . Derivative over original:

\frac{d y}{d x} = 12 e^{3 x} + \frac{2 x}{x ^{2} + 1}

In our answer data, exponentials like $e^{3 x}$ are answered correctly 74% of the time, against 93% for the plain table derivatives. Almost all of the gap is that missing 3.

Answer

\frac{d y}{d x} = 12 e^{3 x} + \frac{2 x}{x ^{2} + 1}

Example 3: A product of both

Harder

Differentiate:

y = e^{2 x} ln (3 x)

This is a product, so the product rule: $f = e^{2 x}$ , $g = ln (3 x)$ .

$f^{'} = 2 e^{2 x}$ by the chain rule. For $g^{'}$ , the inside is $3 x$ , so $g^{'} = \frac{3}{3 x} = \frac{1}{x}$ . The 3 cancels completely, which makes sense: $ln (3 x) = ln 3 + ln x$ , and $ln 3$ is a constant.

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

You can factor out $e^{2 x}$ to write it as $e^{2 x} (2 ln (3 x) + \frac{1}{x})$ . Either form is right.

Answer

\frac{d y}{d x} = 2 e^{2 x} ln (3 x) + \frac{e ^{2 x}}{x}

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Differentiate:

y = 3 e^{- 2 x} + 4 x

Problem 2

Medium

Differentiate:

y = 5 ln (2 x^{3} + 7)

Problem 3

Harder

Differentiate:

y = x^{2} e^{- x^{2}}

Common mistakes

Dropping the inside derivative on $e$ . $\frac{d}{d x} e^{3 x}$ is $3 e^{3 x}$ , not $e^{3 x}$ . The function copies itself, but the chain rule still applies.
Using the power rule on $e^{x}$ . $\frac{d}{d x} e^{x}$ is not $x e^{x - 1}$ . The $x$ is in the exponent, so the power rule has nothing to say.
Writing $\frac{1}{x ^{2} + 1}$ for $\frac{d}{d x} ln (x^{2} + 1)$ . That is one over the inside, but it is missing the inside's derivative on top. The answer is $\frac{2 x}{x ^{2} + 1}$ .
Splitting a log of a sum. The log laws turn products into sums, not the other way round: $ln (x^{2} + 1)$ is not $ln (x^{2}) + ln 1$ . There is no law for $ln (a + b)$ , so a log of a sum has to go through the chain rule as it is.
Changing the exponent. $\frac{d}{d x} e^{x^{2}}$ is $2 x e^{x^{2}}$ . The exponent stays exactly $x^{2}$ ; it does not drop to $x$ or turn into $2 x$ .

Practice it now

Five minutes, no account

Plain $e^{x}$ and $ln x$ (with a number in front) are some of the first problems in a Sprint, at difficulty 1 and 2. Then come $e^{b x}$ and $e^{b x^{2}}$ with a linear term at 2 and 3, and logs of polynomials, $ln (b x^{p} + c)$ , at 3 and 4.

Play a Sprint See it on the formula sheet

FAQ

Why is the derivative of e^x equal to e^x?: That is what makes $e$ special. Every exponential $a^{x}$ has a derivative proportional to itself, $a^{x} ln a$ , and $e \approx 2.718$ is the one base where that constant, $ln e$ , is exactly 1.
What is the derivative of ln(kx)?: Just $\frac{1}{x}$ , for any positive constant $k$ . The chain rule gives $\frac{k}{k x}$ , and the $k$ cancels. Another way to see it: $ln (k x) = ln k + ln x$ , and $ln k$ is a constant.
What are the integrals of e^x and 1/x?: Read the derivatives backwards: $\int e^{x} d x = e^{x} + C$ and $\int \frac{1}{x} d x = ln ∣ x ∣ + C$ . The absolute value is there because $ln ∣ x ∣$ also has derivative $\frac{1}{x}$ for negative $x$ , where plain $ln x$ does not exist. With an inside, divide by its derivative instead of multiplying, as long as it is a constant: $\int e^{3 x} d x = \frac{1}{3} e^{3 x} + C$ .
What about other bases, like 2^x and log base 2?: $\frac{d}{d x} a^{x} = a^{x} ln a$ , and $\frac{d}{d x} lo g_{a} x = \frac{1}{x l n a}$ . With $a = e$ the $ln a$ is 1 and you get back the rules above.
How do you differentiate x^x?: Neither the power rule nor the exponential rule fits, since the variable is in both places. Take logs first: that is logarithmic differentiation, and it gives $x^{x} (ln x + 1)$ .
Where can I find derivatives of e^x and ln x practice problems?: Right on this page: "Your turn" has three derivatives of e^x and ln x practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

When to use it

Use these whenever

e

is raised to a power with

x

in it, or you are taking the natural log of something. The plain versions,

e^{x}

and

ln x

, are straight off the table. Anything else,

e^{3 x}

e^{- x^{2}}

ln (5 x)

ln (x^{2} + 1)

, needs the chain rule on top: the derivative of the inside comes out as a factor. A product like

x^{2} e^{x}

needs the product rule as well. And for a log of a big product or quotient, it is often easiest to split it with the log laws before you differentiate anything.

How to do it, step by step

Find the inside. In

e^{3 x^{2}}

the inside is

3 x^{2}

; in

ln (x^{2} + 1)

it is

x^{2} + 1

. If the inside is just

x

, the table rule is the whole answer.

For $e^{□}$ : copy it, then multiply by the inside's derivative.

e^{3 x^{2}}

stays

e^{3 x^{2}}

and gains a factor of

6 x

For $ln (□)$ : put the inside's derivative over the inside.

ln (x^{2} + 1)

becomes

\frac{2 x}{x ^{2} + 1}

Use the log laws first if they help.

ln (x^{3}) = 3 ln x

and

ln \frac{a}{b} = ln a - ln b

often turn a hard derivative into an easy one.

Worked examples

Example 1: Straight from the table

Easy

Differentiate:

y = 5 e^{x} - 3 ln x

Constants stay in front. $e^{x}$ differentiates to itself, and $ln x$ to $\frac{1}{x}$ .

\frac{d y}{d x} = 5 \cdot e^{x} - 3 \cdot \frac{1}{x}

Answer

\frac{d y}{d x} = 5 e^{x} - \frac{3}{x}

Example 2: With something inside

Medium

Differentiate:

y = 4 e^{3 x} + ln (x^{2} + 1)

First term: the inside is $3 x$ , whose derivative is 3. So $4 e^{3 x}$ becomes $4 \cdot 3 e^{3 x}$ . The exponent itself does not change.

Second term: the inside is $x^{2} + 1$ , whose derivative is $2 x$ . Derivative over original:

\frac{d y}{d x} = 12 e^{3 x} + \frac{2 x}{x ^{2} + 1}

In our answer data, exponentials like $e^{3 x}$ are answered correctly 74% of the time, against 93% for the plain table derivatives. Almost all of the gap is that missing 3.

Answer

\frac{d y}{d x} = 12 e^{3 x} + \frac{2 x}{x ^{2} + 1}

Example 3: A product of both

Harder

Differentiate:

y = e^{2 x} ln (3 x)

This is a product, so the product rule: $f = e^{2 x}$ , $g = ln (3 x)$ .

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

You can factor out $e^{2 x}$ to write it as $e^{2 x} (2 ln (3 x) + \frac{1}{x})$ . Either form is right.

Answer

\frac{d y}{d x} = 2 e^{2 x} ln (3 x) + \frac{e ^{2 x}}{x}

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Differentiate:

y = 3 e^{- 2 x} + 4 x

Problem 2

Medium

Differentiate:

y = 5 ln (2 x^{3} + 7)

Problem 3

Harder

Differentiate:

y = x^{2} e^{- x^{2}}

Common mistakes

Dropping the inside derivative on $e$ .

\frac{d}{d x} e^{3 x}

3 e^{3 x}

, not

e^{3 x}

. The function copies itself, but the chain rule still applies.

Using the power rule on $e^{x}$ .

\frac{d}{d x} e^{x}

is not

x e^{x - 1}

. The

x

is in the exponent, so the power rule has nothing to say.

Writing $\frac{1}{x ^{2} + 1}$ for $\frac{d}{d x} ln (x^{2} + 1)$ . That is one over the inside, but it is missing the inside's derivative on top. The answer is

\frac{2 x}{x ^{2} + 1}

Splitting a log of a sum. The log laws turn products into sums, not the other way round:

ln (x^{2} + 1)

is not

ln (x^{2}) + ln 1

. There is no law for

ln (a + b)

, so a log of a sum has to go through the chain rule as it is.

Changing the exponent.

\frac{d}{d x} e^{x^{2}}

2 x e^{x^{2}}

. The exponent stays exactly

x^{2}

; it does not drop to

x

or turn into

2 x

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

Why is the derivative of e^x equal to e^x?

That is what makes

e

special. Every exponential

a^{x}

has a derivative proportional to itself,

a^{x} ln a

, and

e \approx 2.718

is the one base where that constant,

ln e

, is exactly 1.

What is the derivative of ln(kx)?

Just

\frac{1}{x}

, for any positive constant

k

. The chain rule gives

\frac{k}{k x}

, and the

k

cancels. Another way to see it:

ln (k x) = ln k + ln x

, and

ln k

is a constant.

What are the integrals of e^x and 1/x?

Read the derivatives backwards:

\int e^{x} d x = e^{x} + C

and

\int \frac{1}{x} d x = ln ∣ x ∣ + C

. The absolute value is there because

ln ∣ x ∣

also has derivative

\frac{1}{x}

for negative

x

, where plain

ln x

does not exist. With an inside, divide by its derivative instead of multiplying, as long as it is a constant:

\int e^{3 x} d x = \frac{1}{3} e^{3 x} + C

What about other bases, like 2^x and log base 2?

\frac{d}{d x} a^{x} = a^{x} ln a

, and

\frac{d}{d x} lo g_{a} x = \frac{1}{x l n a}

. With

a = e

the

ln a

is 1 and you get back the rules above.

How do you differentiate x^x?

Neither the power rule nor the exponential rule fits, since the variable is in both places. Take logs first: that is logarithmic differentiation, and it gives

x^{x} (ln x + 1)

Where can I find derivatives of e^x and ln x practice problems?

Right on this page: "Your turn" has three derivatives of e^x and ln x practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

Derivatives of e^x and ln x: Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: Straight from the table

Example 2: With something inside

Example 3: A product of both

Your turn

Common mistakes

FAQ

Keep going

Derivatives of e^x and ln x: Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: Straight from the table

Example 2: With something inside

Example 3: A product of both

Your turn

Common mistakes

FAQ

Keep going