Differentiating ln x and e^x

The exponential function $e^x$ is the one function that is its own derivative, and its inverse $\ln x$ has the simplest possible derivative, $\dfrac{1}{x}$. Together with the chain rule, these two facts let you differentiate every exponential and logarithmic expression on the syllabus. They appear on every P3 paper: inside stationary-point questions, in tangents and normals, in rates of growth and decay, and as the starting point for integrating $\dfrac{1}{x}$ and $e^{x}$.

Why e^x is its own derivative

Look at any exponential curve $y = a^x$ with $a > 1$. It gets steeper as it climbs, and the gradient at each point turns out to be proportional to the height at that point: $\dfrac{d}{dx}a^x = k \cdot a^x$ for some constant $k$ that depends on $a$.

• For $a = 2$, the constant is about $0.693$: the gradient is a bit less than the height.
• For $a = 3$, the constant is about $1.099$: the gradient is a bit more than the height.

Somewhere between 2 and 3 there is a base for which the constant is exactly 1. That base is $e = 2.71828\ldots$, and that is its definition for calculus. For $y = e^x$, the gradient at every point equals the $y$-coordinate. At $(0, 1)$ the gradient is $1$; at $(1, e)$ the gradient is $e$; at $(\ln 5, 5)$ the gradient is $5$.

So differentiating $e^{\text{(something)}}$ leaves the exponential exactly as it was and multiplies it by the derivative of the power.

$$\frac{d}{dx}e^{5x} = 5e^{5x}, \qquad \frac{d}{dx}e^{-x} = -e^{-x}, \qquad \frac{d}{dx}e^{x^2 + 1} = 2x\,e^{x^2+1}, \qquad \frac{d}{dx}e^{\sqrt{x}} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}$$

Why ln x has derivative 1/x

$\ln x$ is the inverse of $e^x$: $y = \ln x$ means exactly $x = e^y$. Differentiate this with respect to $y$:

$$\frac{dx}{dy} = e^y = x$$

and turn it upside down:

$$\frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{x}$$

Geometrically, $y = \ln x$ is the reflection of $y = e^x$ in the line $y = x$. Reflection swaps the roles of $x$ and $y$, so the gradient at a point becomes the reciprocal of the gradient at the mirror-image point. The gradient of $e^x$ at $(0, 1)$ is $1$, so the gradient of $\ln x$ at $(1, 0)$ is $1$; the gradient of $e^x$ at $(\ln 4, 4)$ is $4$, so the gradient of $\ln x$ at $(4, \ln 4)$ is $\tfrac{1}{4}$.

The curves $y = e^x$ (solid) and $y = \ln x$ (dashed) are reflections of each other in $y = x$. Both have gradient $1$ where they cross the axes.

The general form is worth reading aloud: the derivative of $\ln$ of something is the derivative of the something, divided by the something.

$$\frac{d}{dx}\ln(x^2 - 3x) = \frac{2x - 3}{x^2 - 3x}, \qquad \frac{d}{dx}\ln(1 + e^x) = \frac{e^x}{1 + e^x}, \qquad \frac{d}{dx}\ln(\cos x) = \frac{-\sin x}{\cos x} = -\tan x$$

The surprising case ln(kx)

$$\frac{d}{dx}\ln 4x = \frac{4}{4x} = \frac{1}{x}$$

This looks wrong at first, but it is right: $\ln 4x = \ln 4 + \ln x$, and $\ln 4$ is a constant, which differentiates to zero. The graphs of $\ln x$, $\ln 4x$ and $\ln 100x$ are vertical translations of one another, so they have the same gradient at every $x$.

Use the log laws before you differentiate

The laws of logarithms turn products, quotients and powers inside a log into sums, differences and multiples outside it. Since sums are much easier to differentiate than quotients, always expand first.

$$y = \ln\frac{x^2}{\sqrt{3x+1}} = 2\ln x - \tfrac{1}{2}\ln(3x+1) \quad\Rightarrow\quad \frac{dy}{dx} = \frac{2}{x} - \frac{3}{2(3x+1)}$$

Without the expansion you would need the quotient rule and the chain rule nested inside the chain rule.

Also distinguish:

Expression	Means	Derivative
$\ln x^3$	$\ln(x^3) = 3\ln x$	$\dfrac{3}{x}$
$(\ln x)^3$	the cube of $\ln x$	$3(\ln x)^2 \cdot \dfrac{1}{x} = \dfrac{3(\ln x)^2}{x}$
$\ln(\ln x)$	log of a log	$\dfrac{1}{\ln x}\cdot\dfrac{1}{x} = \dfrac{1}{x\ln x}$

Simplifying exponentials first

Index laws do the same job for exponentials.

• $\dfrac{2}{e^{3x}} = 2e^{-3x}$, derivative $-6e^{-3x}$.
• $(e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x}$, derivative $2e^{2x} - 2e^{-2x}$.
• $\dfrac{e^{2x} - 1}{e^x} = e^{x} - e^{-x}$, derivative $e^x + e^{-x}$.
• $e^{\ln x} = x$ and $e^{2\ln x} = x^2$: simplify these on sight.

Equations that come out of these derivatives

Setting a derivative to zero frequently gives an equation in $e^x$ or $\ln x$. These standard moves finish the job:

• $e^{kx} = c \Rightarrow kx = \ln c$, valid only for $c > 0$. If $c \le 0$ there is no solution, and that is often the point of the question.
• $\ln x = c \Rightarrow x = e^c$.
• An equation in $e^{2x}$ and $e^x$ is a quadratic in $u = e^x$, because $e^{2x} = (e^x)^2$.
• $4e^x = 9e^{-x}$: multiply through by $e^x$ to get $4e^{2x} = 9$.

See logarithms and Euler's number for more on solving them.

Worked examples

Practice

Answers

1. (a) $5 \times (-3)e^{2-3x} = -15e^{2-3x}$. (b) The power $x^{1/2}$ has derivative $\tfrac{1}{2}x^{-1/2}$: $\dfrac{e^{\sqrt{x}}}{2\sqrt{x}}$. (c) Expand: $e^{2x} + 2 + e^{-2x}$, derivative $2e^{2x} - 2e^{-2x}$.

2. (a) $\dfrac{-2x}{7 - x^2}$. (b) $y = 2\ln(x+1) - \ln(x-2)$, so $\dfrac{dy}{dx} = \dfrac{2}{x + 1} - \dfrac{1}{x - 2}$, which simplifies to $\dfrac{x - 5}{(x+1)(x-2)}$. (c) $\dfrac{1}{\ln x}\cdot\dfrac{1}{x} = \dfrac{1}{x\ln x}$.

3. At $x = 2$: $y = e^0 = 1$, $\dfrac{dy}{dx} = 2e^{2x-4} = 2$. Tangent: $y - 1 = 2(x - 2)$, i.e. $y = 2x - 3$.

4. $\dfrac{dy}{dx} = 3e^{3x} - 6 = 0 \Rightarrow e^{3x} = 2 \Rightarrow x = \tfrac{1}{3}\ln 2$. Then $y = 2 - 6 \cdot \tfrac{1}{3}\ln 2 = 2 - 2\ln 2$. $\dfrac{d^2y}{dx^2} = 9e^{3x} = 18 > 0$, a minimum at $\left(\tfrac{1}{3}\ln 2,\ 2 - 2\ln 2\right)$.

5. $\dfrac{dy}{dx} = 1 - \dfrac{2}{x + 1} = 0 \Rightarrow x = 1$, $y = 1 - 2\ln 2$. $\dfrac{d^2y}{dx^2} = \dfrac{2}{(x+1)^2} = \tfrac{1}{2} > 0$: a minimum at $(1,\ 1 - 2\ln 2)$.

6. At $x = a$ the tangent is $y - 3\ln a = \dfrac{3}{a}(x - a)$. Through $(0, 0)$: $-3\ln a = -3$, so $\ln a = 1$, $a = e$. The point is $(e, 3)$.

7. $(0, 5)$ gives $A = 5$. $\dfrac{dy}{dx} = 5ke^{kx}$, and at $x = 0$ this is $5k = -10$, so $k = -2$. Then $-10e^{-2x} = -0.1 \Rightarrow e^{-2x} = 0.01 \Rightarrow -2x = \ln 0.01 = -2\ln 10$, so $x = \ln 10$.

8. $\dfrac{dy}{dx} = 2e^{2x} - 2e^x + 2 = 2(u^2 - u + 1)$ with $u = e^x$. Since $u^2 - u + 1 = \left(u - \tfrac{1}{2}\right)^2 + \tfrac{3}{4} > 0$, the derivative is always positive and never zero. No stationary points.

9. $\dfrac{dy}{dx} = \dfrac{2}{2x - 1} - \dfrac{a}{x^2}$. At $x = 2$: $\tfrac{2}{3} - \tfrac{a}{4} = 0 \Rightarrow a = \tfrac{8}{3}$. Then $y = \ln 3 + \dfrac{8/3}{2} = \ln 3 + \tfrac{4}{3}$. $\dfrac{d^2y}{dx^2} = -\dfrac{4}{(2x-1)^2} + \dfrac{2a}{x^3}$; at $x = 2$: $-\tfrac{4}{9} + \dfrac{16/3}{8} = -\tfrac{4}{9} + \tfrac{2}{3} = \tfrac{2}{9} > 0$, so a minimum at $\left(2,\ \ln 3 + \tfrac{4}{3}\right)$.