Mathematics 9709 · Pure Mathematics 2

Differentiating e^x, a^x and ln x

Two derivatives to memorise, then one habit.
$e^x$ is the function that is its own gradient; $\ln x$ differentiates to $\tfrac1x$ .
Everything else is just the chain rule: multiply by the derivative of the inside.

y = eˣ is its own gradient; y = ln x (its mirror in y = x) has gradient 1/x.

The two results to memorise

$\dfrac{d}{dx}(e^x)=e^x$ and $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ . (Here $\ln$ means natural log, base $e$ . It always does, in 9709.)

$e^x$ being its own gradient is the whole reason the number $e$ matters. (Key Points 4.3, 4.5)

Why d/dx(ln x) = 1/x

Start from $y=\ln x$ , which means $x=e^y$ . Differentiate that with respect to $y$ : $\dfrac{dx}{dy}=e^y=x$ . Now flip it: $\dfrac{dy}{dx}=\dfrac{1}{\,dx/dy\,}=\dfrac{1}{x}$ . So the log rule is just the $e^x$ rule read backwards. (textbook §4.4)

The chain-rule forms (what papers test)

A plain $e^x$ or $\ln x$ almost never shows up alone; what you get is something inside:

$\dfrac{d}{dx}\,e^{\,\mathrm{f}(x)}=\mathrm{f}'(x)\,e^{\,\mathrm{f}(x)}$ : keep the $e$ , then multiply by the derivative of the inside. And $\dfrac{d}{dx}\ln \mathrm{f}(x)=\dfrac{\mathrm{f}'(x)}{\mathrm{f}(x)}$ : the derivative of the inside goes on top.

Function	Inside derivative	Answer
$e^{2x}$	$2$	$2e^{2x}$
$e^{3x^2-1}$	$6x$	$6x\,e^{3x^2-1}$
$\ln(2x-1)$	$2$	$\dfrac{2}{2x-1}$
$\ln(x^2+1)$	$2x$	$\dfrac{2x}{x^2+1}$

In particular $\frac{d}{dx}e^{ax+b}=a\,e^{ax+b}$ and $\frac{d}{dx}\ln(ax+b)=\frac{a}{ax+b}$ . (Key Points 4.4, 4.6)

Common mistake

The biggest error in this topic: writing

\tfrac{d}{dx}e^{3x}=e^{3x}

(forgetting the

\times 3

) or

\tfrac{d}{dx}\ln(2x-1)=\tfrac{1}{2x-1}

(forgetting the

\times 2

on top). The inside derivative is the method mark, so never leave it off.

Why d/dx(ln 3x) = 1/x

By the rule, $\tfrac{d}{dx}\ln 3x=\tfrac{3}{3x}=\tfrac1x$ , the same as $\tfrac{d}{dx}\ln 7x$ . The number out front drops out because $\ln 3x=\ln 3+\ln x$ and $\ln 3$ is a constant. Spotting this saves you a chain-rule step. (textbook Ex4D)

Use ln-laws first

Before you differentiate a complicated log, simplify it with ln-laws first. A power moves to the front. A product or quotient splits into a sum or a difference. A long quotient-rule problem turns into a one-line sum.

Worked example

Differentiate $y = \ln\sqrt{5-x}$

Bring the power down first: $\ln\sqrt{5-x}=\tfrac12\ln(5-x)$ .
Now it is a simple chain-rule log (inside derivative $-1$ ): $\dfrac{dy}{dx}=\tfrac12\cdot\dfrac{-1}{5-x}$ .

Answer

\dfrac{dy}{dx}=\dfrac{-1}{2(5-x)}

No quotient rule, no product rule needed. (textbook Worked Example 4.7)

Worked example

Differentiate $y = \ln\!\big(x^2(2x+1)^3\big)$ : split it before you touch it

Split the product, then drop each power: $\ln\!\big(x^2(2x+1)^3\big)=2\ln x+3\ln(2x+1)$ .
Differentiate each piece (second is a chain-rule log, inside derivative $2$ ): $\dfrac{dy}{dx}=\dfrac{2}{x}+3\cdot\dfrac{2}{2x+1}$ .

Answer

\dfrac{dy}{dx}=\dfrac{2}{x}+\dfrac{6}{2x+1}

The other way, a product rule inside a chain-rule log, is far more work and far more likely to go wrong. (textbook Worked Example 4.8)

And a^x: rewrite it as e to a power

For a base that is not $e$ , turn it into one: $a^x=e^{x\ln a}$ , then use the chain rule (inside derivative $\ln a$ ).

$\dfrac{d}{dx}(a^x)=a^x\ln a$ . The $\ln a$ is part of the exact answer, so keep it (e.g. $2^x\ln 2$ ).

This is rare, so one idea is enough. Underneath, everything is just $e$ . (textbook Ex4C)

Where it shows up

Built into bigger questions, not on their own: an exp or log with the product or quotient rule (§4.0), then a stationary point with exact coordinates, or a tangent/normal.
The same e-derivatives also turn up in implicit (§4.3) and parametric (§4.4) questions.

Your turn— tap to reveal the worked answer (9709-style: y = x² ln x, exact stationary point)

Product rule: $\tfrac{dy}{dx}=x^2\cdot\tfrac1x+\ln x\cdot 2x=x(1+2\ln x)$ . Zero when $1+2\ln x=0$ , i.e. $\ln x=-\tfrac12$ :

Answer

x=e^{-1/2},\quad y=-\dfrac{1}{2e}

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Differentiating e^x, a^x and ln x

Two derivatives to memorise, then one habit.
$e^x$ is the function that is its own gradient; $\ln x$ differentiates to $\tfrac1x$ .
Everything else is just the chain rule: multiply by the derivative of the inside.

y = eˣ is its own gradient; y = ln x (its mirror in y = x) has gradient 1/x.

The two results to memorise

$\dfrac{d}{dx}(e^x)=e^x$ and $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ . (Here $\ln$ means natural log, base $e$ . It always does, in 9709.)

$e^x$ being its own gradient is the whole reason the number $e$ matters. (Key Points 4.3, 4.5)

Why d/dx(ln x) = 1/x

The chain-rule forms (what papers test)

A plain $e^x$ or $\ln x$ almost never shows up alone; what you get is something inside:

Function	Inside derivative	Answer
$e^{2x}$	$2$	$2e^{2x}$
$e^{3x^2-1}$	$6x$	$6x\,e^{3x^2-1}$
$\ln(2x-1)$	$2$	$\dfrac{2}{2x-1}$
$\ln(x^2+1)$	$2x$	$\dfrac{2x}{x^2+1}$

In particular $\frac{d}{dx}e^{ax+b}=a\,e^{ax+b}$ and $\frac{d}{dx}\ln(ax+b)=\frac{a}{ax+b}$ . (Key Points 4.4, 4.6)

Common mistake

The biggest error in this topic: writing

\tfrac{d}{dx}e^{3x}=e^{3x}

(forgetting the

\times 3

) or

\tfrac{d}{dx}\ln(2x-1)=\tfrac{1}{2x-1}

(forgetting the

\times 2

on top). The inside derivative is the method mark, so never leave it off.

Why d/dx(ln 3x) = 1/x

Use ln-laws first

Worked example

Differentiate $y = \ln\sqrt{5-x}$

Bring the power down first: $\ln\sqrt{5-x}=\tfrac12\ln(5-x)$ .
Now it is a simple chain-rule log (inside derivative $-1$ ): $\dfrac{dy}{dx}=\tfrac12\cdot\dfrac{-1}{5-x}$ .

Answer

\dfrac{dy}{dx}=\dfrac{-1}{2(5-x)}

No quotient rule, no product rule needed. (textbook Worked Example 4.7)

Worked example

Differentiate $y = \ln\!\big(x^2(2x+1)^3\big)$ : split it before you touch it

Split the product, then drop each power: $\ln\!\big(x^2(2x+1)^3\big)=2\ln x+3\ln(2x+1)$ .
Differentiate each piece (second is a chain-rule log, inside derivative $2$ ): $\dfrac{dy}{dx}=\dfrac{2}{x}+3\cdot\dfrac{2}{2x+1}$ .

Answer

\dfrac{dy}{dx}=\dfrac{2}{x}+\dfrac{6}{2x+1}

The other way, a product rule inside a chain-rule log, is far more work and far more likely to go wrong. (textbook Worked Example 4.8)

And a^x: rewrite it as e to a power

For a base that is not $e$ , turn it into one: $a^x=e^{x\ln a}$ , then use the chain rule (inside derivative $\ln a$ ).

$\dfrac{d}{dx}(a^x)=a^x\ln a$ . The $\ln a$ is part of the exact answer, so keep it (e.g. $2^x\ln 2$ ).

This is rare, so one idea is enough. Underneath, everything is just $e$ . (textbook Ex4C)

Where it shows up

Built into bigger questions, not on their own: an exp or log with the product or quotient rule (§4.0), then a stationary point with exact coordinates, or a tangent/normal.
The same e-derivatives also turn up in implicit (§4.3) and parametric (§4.4) questions.

Your turn— tap to reveal the worked answer (9709-style: y = x² ln x, exact stationary point)

Product rule: $\tfrac{dy}{dx}=x^2\cdot\tfrac1x+\ln x\cdot 2x=x(1+2\ln x)$ . Zero when $1+2\ln x=0$ , i.e. $\ln x=-\tfrac12$ :

Answer

x=e^{-1/2},\quad y=-\dfrac{1}{2e}