Topics

Differentiating exponential and logarithmic functions

Covers AS91578

Base $e$ only

The standard restricts logarithms and exponentials to base $e$ . You will not be asked to differentiate $\log_{10} x$ or $10^x$ .
$e \approx 2.71828$ is the number for which the exponential function is its own derivative. That property is why calculus uses it.
$\ln x$ means $\log_e x$ — the natural logarithm.

The two standard results

$\frac{d}{dx}\left(e^x\right) = e^x$

$\frac{d}{dx}\left(\ln x\right) = \frac{1}{x}$

$e^x$ is unchanged by differentiation. No other non-zero function does this.
$\ln x$ differentiates to a power function, which is the reason logarithms appear inside integration questions too.

With a linear function inside

These are the versions you actually use, and they follow from the chain rule:

$\frac{d}{dx}\left(e^{ax+b}\right) = a\,e^{ax+b}$

$\frac{d}{dx}\left(\ln(ax+b)\right) = \frac{a}{ax+b}$

The multiplier $a$ is the derivative of the inside. Forgetting it is the single most common error with these functions.
Worked through:
- $\dfrac{d}{dx}\left(e^{5x}\right) = 5e^{5x}$
- $\dfrac{d}{dx}\left(e^{-2x+7}\right) = -2e^{-2x+7}$
- $\dfrac{d}{dx}\left(\ln(3x - 1)\right) = \dfrac{3}{3x-1}$

The general forms

For any inner function $u$ :

$\frac{d}{dx}\left(e^{u}\right) = u'\,e^{u} \qquad \frac{d}{dx}\left(\ln u\right) = \frac{u'}{u}$

$\dfrac{d}{dx}\left(e^{x^2}\right) = 2x\,e^{x^2}$
$\dfrac{d}{dx}\left(\ln(x^2 + 1)\right) = \dfrac{2x}{x^2+1}$
Note the shape of the log derivative: it is always derivative of the inside over the inside. Recognising that shape saves time and prevents the quotient rule being used unnecessarily.

Using log laws to simplify first

Differentiating a complicated logarithm is far easier after the log laws have broken it up.
The three laws:
- $\ln(AB) = \ln A + \ln B$
- $\ln\left(\dfrac{A}{B}\right) = \ln A - \ln B$
- $\ln\left(A^n\right) = n\ln A$
Worked through — differentiate $y = \ln\left(x^3\sqrt{2x+1}\right)$ :
- Expand first: $y = 3\ln x + \frac{1}{2}\ln(2x+1)$
- Differentiate each simple piece: $\dfrac{dy}{dx} = \dfrac{3}{x} + \dfrac{1}{2} \times \dfrac{2}{2x+1} = \dfrac{3}{x} + \dfrac{1}{2x+1}$
Doing that directly would have needed the chain rule twice and the product rule once. Simplify first is worth real marks.

Curve features of these functions

$y = e^x$ is always positive, always increasing, and always concave up. It has a horizontal asymptote $y = 0$ as $x \to -\infty$ .
$y = \ln x$ exists only for $x > 0$ , is always increasing, and is always concave down. It has a vertical asymptote at $x = 0$ .
Note the domain restriction on $\ln$ . Any solution to an equation that requires $\ln$ of a negative number or zero must be rejected.

Solving the equations that come out of these derivatives

Setting $f'(x) = 0$ $f^{'} (x) = 0$ with exponentials often produces an equation needing logs:
- $e^{2x} = 5 \;\Rightarrow\; 2x = \ln 5 \;\Rightarrow\; x = \frac{\ln 5}{2}$
A key fact: $e^{\text{anything}}$ is never zero. So in $f'(x) = (2x - 6)e^x = 0$ , only $2x - 6 = 0$ can give a solution, and $x = 3$ is the only stationary point.

Tips and Tricks

The multiplier is the whole game. $\dfrac{d}{dx}\left(e^{4x}\right)$ is $4e^{4x}$ , not $e^{4x}$ . The 2025 report notes candidates who "frequently omitted a factor of the differentiation, which led to an incorrect differentiation, so no credit could be given for that question part" — even for those who went on to solve the rest correctly.

Use the log laws before differentiating, not after. Expanding $\ln$ of a product, quotient or power into a sum turns one hard derivative into two or three easy ones.

$\dfrac{d}{dx}(\ln u) = \dfrac{u'}{u}$ — the derivative goes on top. Writing $\dfrac{u}{u'}$ or $\dfrac{1}{u'}$ are both common and both wrong.

$e^{\text{something}}$ is never zero. When you factorise a derivative and one factor is an exponential, that factor cannot contribute a stationary point. Setting it to zero and "solving" is a wasted line — say instead that it is always positive, so only the other factor matters.

Check the domain on any answer involving $\ln$ . If a stationary point comes out at $x = -2$ for a function containing $\ln x$ , it must be rejected because $\ln(-2)$ does not exist.

Worked ExampleCombining log laws with differentiation

Find $\dfrac{dy}{dx}$ for $y = \ln\left(\dfrac{(x+2)^4}{\sqrt{3x - 1}}\right)$ .

Step 1 — Recognise the opportunity

Differentiating this directly would need the chain rule around a quotient rule. Instead, expand the logarithm first using the log laws, which turns it into a sum of three simple derivatives.

Step 2 — Split the quotient

Apply $\ln\left(\frac{A}{B}\right) = \ln A - \ln B$ :

$y = \ln\left((x+2)^4\right) - \ln\left(\sqrt{3x-1}\right)$

Step 3 — Bring down the powers

Apply $\ln(A^n) = n\ln A$ to both terms, writing the square root as a power of $\frac{1}{2}$ :

$y = 4\ln(x+2) - \frac{1}{2}\ln(3x - 1)$

Step 4 — Differentiate the first term

Use $\dfrac{d}{dx}\ln u = \dfrac{u'}{u}$ , where $u = x + 2$ so $u' = 1$ :

$\frac{d}{dx}\left[4\ln(x+2)\right] = 4 \times \frac{1}{x+2} = \frac{4}{x+2}$

Step 5 — Differentiate the second term

Here $u = 3x - 1$ , so $u' = 3$ :

$\frac{d}{dx}\left[-\frac{1}{2}\ln(3x-1)\right] = -\frac{1}{2} \times \frac{3}{3x - 1} = -\frac{3}{2(3x-1)}$

Note the multiplier 3 — it comes from the derivative of the inside, and dropping it is the standard error here.

Step 6 — Combine

$\frac{dy}{dx} = \frac{4}{x+2} - \frac{3}{2(3x-1)}$

Step 7 — Sense-check the domain

The original function needs $x + 2 > 0$ and $3x - 1 > 0$ , so $x > \frac{1}{3}$ . The derivative is defined on that whole interval, with no zeros of either denominator inside it ✓

Differentiating exponential and logarithmic functions

Covers AS91578

Base $e$ only

The standard restricts logarithms and exponentials to base $e$ . You will not be asked to differentiate $\log_{10} x$ or $10^x$ .
$e \approx 2.71828$ is the number for which the exponential function is its own derivative. That property is why calculus uses it.
$\ln x$ means $\log_e x$ — the natural logarithm.

The two standard results

$\frac{d}{dx}\left(e^x\right) = e^x$

$\frac{d}{dx}\left(\ln x\right) = \frac{1}{x}$

$e^x$ is unchanged by differentiation. No other non-zero function does this.
$\ln x$ differentiates to a power function, which is the reason logarithms appear inside integration questions too.

With a linear function inside

These are the versions you actually use, and they follow from the chain rule:

$\frac{d}{dx}\left(e^{ax+b}\right) = a\,e^{ax+b}$

$\frac{d}{dx}\left(\ln(ax+b)\right) = \frac{a}{ax+b}$

The multiplier $a$ is the derivative of the inside. Forgetting it is the single most common error with these functions.
Worked through:
- $\dfrac{d}{dx}\left(e^{5x}\right) = 5e^{5x}$
- $\dfrac{d}{dx}\left(e^{-2x+7}\right) = -2e^{-2x+7}$
- $\dfrac{d}{dx}\left(\ln(3x - 1)\right) = \dfrac{3}{3x-1}$

The general forms

For any inner function $u$ :

$\frac{d}{dx}\left(e^{u}\right) = u'\,e^{u} \qquad \frac{d}{dx}\left(\ln u\right) = \frac{u'}{u}$

$\dfrac{d}{dx}\left(e^{x^2}\right) = 2x\,e^{x^2}$
$\dfrac{d}{dx}\left(\ln(x^2 + 1)\right) = \dfrac{2x}{x^2+1}$
Note the shape of the log derivative: it is always derivative of the inside over the inside. Recognising that shape saves time and prevents the quotient rule being used unnecessarily.

Using log laws to simplify first

Differentiating a complicated logarithm is far easier after the log laws have broken it up.
The three laws:
- $\ln(AB) = \ln A + \ln B$
- $\ln\left(\dfrac{A}{B}\right) = \ln A - \ln B$
- $\ln\left(A^n\right) = n\ln A$
Worked through — differentiate $y = \ln\left(x^3\sqrt{2x+1}\right)$ :
- Expand first: $y = 3\ln x + \frac{1}{2}\ln(2x+1)$
- Differentiate each simple piece: $\dfrac{dy}{dx} = \dfrac{3}{x} + \dfrac{1}{2} \times \dfrac{2}{2x+1} = \dfrac{3}{x} + \dfrac{1}{2x+1}$
Doing that directly would have needed the chain rule twice and the product rule once. Simplify first is worth real marks.

Curve features of these functions

$y = e^x$ is always positive, always increasing, and always concave up. It has a horizontal asymptote $y = 0$ as $x \to -\infty$ .
$y = \ln x$ exists only for $x > 0$ , is always increasing, and is always concave down. It has a vertical asymptote at $x = 0$ .
Note the domain restriction on $\ln$ . Any solution to an equation that requires $\ln$ of a negative number or zero must be rejected.

Solving the equations that come out of these derivatives

Setting $f'(x) = 0$ $f^{'} (x) = 0$ with exponentials often produces an equation needing logs:
- $e^{2x} = 5 \;\Rightarrow\; 2x = \ln 5 \;\Rightarrow\; x = \frac{\ln 5}{2}$
A key fact: $e^{\text{anything}}$ is never zero. So in $f'(x) = (2x - 6)e^x = 0$ , only $2x - 6 = 0$ can give a solution, and $x = 3$ is the only stationary point.

Tips and Tricks

Use the log laws before differentiating, not after. Expanding $\ln$ of a product, quotient or power into a sum turns one hard derivative into two or three easy ones.

$\dfrac{d}{dx}(\ln u) = \dfrac{u'}{u}$ — the derivative goes on top. Writing $\dfrac{u}{u'}$ or $\dfrac{1}{u'}$ are both common and both wrong.

Check the domain on any answer involving $\ln$ . If a stationary point comes out at $x = -2$ for a function containing $\ln x$ , it must be rejected because $\ln(-2)$ does not exist.

Worked ExampleCombining log laws with differentiation

Find $\dfrac{dy}{dx}$ for $y = \ln\left(\dfrac{(x+2)^4}{\sqrt{3x - 1}}\right)$ .

Step 1 — Recognise the opportunity

Differentiating this directly would need the chain rule around a quotient rule. Instead, expand the logarithm first using the log laws, which turns it into a sum of three simple derivatives.

Step 2 — Split the quotient

Apply $\ln\left(\frac{A}{B}\right) = \ln A - \ln B$ :

$y = \ln\left((x+2)^4\right) - \ln\left(\sqrt{3x-1}\right)$

Step 3 — Bring down the powers

Apply $\ln(A^n) = n\ln A$ to both terms, writing the square root as a power of $\frac{1}{2}$ :

$y = 4\ln(x+2) - \frac{1}{2}\ln(3x - 1)$

Step 4 — Differentiate the first term

Use $\dfrac{d}{dx}\ln u = \dfrac{u'}{u}$ , where $u = x + 2$ so $u' = 1$ :

$\frac{d}{dx}\left[4\ln(x+2)\right] = 4 \times \frac{1}{x+2} = \frac{4}{x+2}$

Step 5 — Differentiate the second term

Here $u = 3x - 1$ , so $u' = 3$ :

$\frac{d}{dx}\left[-\frac{1}{2}\ln(3x-1)\right] = -\frac{1}{2} \times \frac{3}{3x - 1} = -\frac{3}{2(3x-1)}$

Note the multiplier 3 — it comes from the derivative of the inside, and dropping it is the standard error here.

Step 6 — Combine

$\frac{dy}{dx} = \frac{4}{x+2} - \frac{3}{2(3x-1)}$

Step 7 — Sense-check the domain

The original function needs $x + 2 > 0$ and $3x - 1 > 0$ , so $x > \frac{1}{3}$ . The derivative is defined on that whole interval, with no zeros of either denominator inside it ✓

Differentiating exponential and logarithmic functions

Base eee only

The two standard results

With a linear function inside

The general forms

Using log laws to simplify first

Curve features of these functions

Solving the equations that come out of these derivatives

Step 1 — Recognise the opportunity

Step 2 — Split the quotient

Step 3 — Bring down the powers

Step 4 — Differentiate the first term

Step 5 — Differentiate the second term

Step 6 — Combine

Step 7 — Sense-check the domain

Differentiating exponential and logarithmic functions

Base eee only

The two standard results

With a linear function inside

The general forms

Using log laws to simplify first

Curve features of these functions

Solving the equations that come out of these derivatives

Step 1 — Recognise the opportunity

Step 2 — Split the quotient

Step 3 — Bring down the powers

Step 4 — Differentiate the first term

Step 5 — Differentiate the second term

Step 6 — Combine

Step 7 — Sense-check the domain

Base $e$ only

Base $e$ only