# Exponential and Log Derivatives Practice Problems with Answers

> 12 practice problems on derivatives of exponential and logarithmic functions, with answers and full solutions: e^x, a^x, ln x, log rules and x^x.

Source: https://practicecalculus.com/practice-problems/derivatives-of-exponential-and-logarithmic-functions
Updated: 2026-11-06

$$ \frac{d}{dx}e^x = e^x, \qquad \frac{d}{dx}\ln x = \frac{1}{x} $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 2 and Unit 3
- **You need:** Log rules and the [chain rule](/practice-problems/chain-rule)
- **Main idea:** e^u differentiates to e^u times u', and ln u differentiates to u' over u
- **Practice with new numbers:** [Exponential and log derivatives lesson](/learn/calculus-1/derivative-rules/exponential-and-log-derivatives)

## Key takeaways

- $e^x$ is its own derivative, and any other base picks up a factor of its natural log: $(a^x)' = a^x\ln a$.
- Use log rules to break a logarithm apart before you differentiate. It turns quotients and roots into simple sums.
- When the variable is in both the base and the exponent, as in $x^x$, take $\ln$ of both sides first.

These **derivatives of exponential and logarithmic functions practice problems** start with the basic rules, then use the chain rule, log properties and logarithmic differentiation. Try each one first. The answers come right after the list, then the full solutions.

The rules you need:

| Function | Derivative |
| --- | --- |
| $e^x$ | $e^x$ |
| $a^x$ | $a^x\ln a$ |
| $\ln x$ | $\dfrac{1}{x}$ |
| $\log_a x$ | $\dfrac{1}{x\ln a}$ |

With an inside function $u$: $(e^u)' = e^u u'$ and $(\ln u)' = \frac{u'}{u}$. See the [derivative of e^x](/derivatives/e-to-the-x), the [derivative of ln x](/derivatives/ln-x) and the [derivative of 2^x](/derivatives/2-to-the-x) for the proofs.

## Problems

**Basic rules**

1. $y = 5e^x - \ln x$
2. $y = 2^x$
3. $y = \log_{10} x$

**Chain rule**

4. $y = e^{3x}$
5. $y = \ln(x^2 + 1)$
6. $y = 5^{x^2}$

**Use log rules first**

7. $y = \ln(5x)$
8. $y = \ln\sqrt{x^2 + 4}$
9. $y = \ln\!\left(\dfrac{x + 1}{x - 1}\right)$

**With other rules**

10. $y = x^2\ln x$
11. $y = \dfrac{e^x}{1 + e^x}$
12. $y = x^x$ for $x > 0$

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $5e^x - \dfrac{1}{x}$ |
| 2 | $2^x\ln 2$ |
| 3 | $\dfrac{1}{x\ln 10}$ |
| 4 | $3e^{3x}$ |
| 5 | $\dfrac{2x}{x^2 + 1}$ |
| 6 | $2x\ln 5\cdot 5^{x^2}$ |
| 7 | $\dfrac{1}{x}$ |
| 8 | $\dfrac{x}{x^2 + 4}$ |
| 9 | $-\dfrac{2}{x^2 - 1}$ |
| 10 | $2x\ln x + x$ |
| 11 | $\dfrac{e^x}{(1 + e^x)^2}$ |
| 12 | $x^x(\ln x + 1)$ |

## Step-by-step solutions

**1.** Differentiate each term:

$$y' = 5e^x - \frac{1}{x}$$

**2.** Use $(a^x)' = a^x\ln a$ with $a = 2$:

$$y' = 2^x\ln 2$$

**3.** Use $(\log_a x)' = \frac{1}{x\ln a}$ with $a = 10$, or write $\log_{10} x = \frac{\ln x}{\ln 10}$ first:

$$y' = \frac{1}{x\ln 10}$$

**4.** The inside is $3x$, with derivative $3$:

$$y' = e^{3x}\cdot 3 = 3e^{3x}$$

**5.** Use $(\ln u)' = \frac{u'}{u}$ with $u = x^2 + 1$:

$$y' = \frac{2x}{x^2 + 1}$$

**6.** Use $(a^u)' = a^u\ln a\cdot u'$ with $a = 5$ and $u = x^2$:

$$y' = 5^{x^2}\ln 5\cdot 2x = 2x\ln 5\cdot 5^{x^2}$$

**7.** Split the log: $\ln(5x) = \ln 5 + \ln x$. The first term is a constant:

$$y' = 0 + \frac{1}{x} = \frac{1}{x}$$

The chain rule gives the same thing: $\frac{5}{5x} = \frac{1}{x}$.

**8.** A square root is a power of $\frac{1}{2}$, and logs bring powers to the front:

$$y = \frac{1}{2}\ln(x^2 + 4) \quad\Rightarrow\quad y' = \frac{1}{2}\cdot\frac{2x}{x^2 + 4} = \frac{x}{x^2 + 4}$$

**9.** Split the quotient: $y = \ln(x + 1) - \ln(x - 1)$. Then

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

This is much shorter than the quotient rule inside the chain rule.

**10.** Use the [product rule](/rules/product-rule):

$$y' = 2x\ln x + x^2\cdot\frac{1}{x} = 2x\ln x + x$$

**11.** Use the [quotient rule](/rules/quotient-rule). Both the top and the bottom have derivative $e^x$:

$$y' = \frac{e^x(1 + e^x) - e^x\cdot e^x}{(1 + e^x)^2} = \frac{e^x + e^{2x} - e^{2x}}{(1 + e^x)^2} = \frac{e^x}{(1 + e^x)^2}$$

**12.** Neither the power rule nor the exponential rule fits, because $x$ is in both places. Use [logarithmic differentiation](/rules/logarithmic-differentiation): take $\ln$ of both sides.

$$\ln y = x\ln x$$

Differentiate both sides, using the product rule on the right:

$$\frac{y'}{y} = \ln x + x\cdot\frac{1}{x} = \ln x + 1 \quad\Rightarrow\quad y' = x^x(\ln x + 1)$$

## Common mistakes on these problems

- **Using the power rule on $e^x$ or $2^x$.** $(2^x)'$ is not $x\cdot 2^{x-1}$. The variable is in the exponent, so you need $2^x\ln 2$.
- **Writing $\ln(x^2 + 1)' = \frac{1}{x^2 + 1}$.** That misses the inside derivative $2x$ on top.
- **Splitting a log the wrong way.** $\ln(x + 1) \ne \ln x + \ln 1$. Log rules split products and quotients, not sums.
- **Treating $x^x$ as $x\cdot x^{x-1}$.** Problem 12 needs logarithmic differentiation. See the [derivative of x^x](/derivatives/x-to-the-x) for more.

## Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new functions and instant feedback, try the [exponential and log derivatives lesson](/learn/calculus-1/derivative-rules/exponential-and-log-derivatives), then the [chain rule with exponentials lesson](/learn/calculus-1/derivative-rules/chain-rule-exponentials). For more worked examples, read the [exponential and log derivatives guide](/blog/derivatives-of-exponential-and-logarithmic-functions).

## Frequently asked questions

### What is the derivative of e to the x?

$\frac{d}{dx}e^x = e^x$. With something inside, use the chain rule: $\frac{d}{dx}e^{u} = e^{u}u'$, so $\frac{d}{dx}e^{3x} = 3e^{3x}$.

### What is the derivative of ln x?

$\frac{d}{dx}\ln x = \frac{1}{x}$ for $x > 0$. With an inside function, $\frac{d}{dx}\ln u = \frac{u'}{u}$. Notice that $\ln(5x)$ also has derivative $\frac{1}{x}$, because $\ln(5x) = \ln 5 + \ln x$.

### What is the derivative of a to the x?

$\frac{d}{dx}a^x = a^x\ln a$ for any positive base $a$. For example, $\frac{d}{dx}2^x = 2^x\ln 2$. When $a = e$, $\ln e = 1$ and you get back $e^x$.