Derivatives of exponentials and logarithms

$e^x$ is its own slope

At every point, the slope of $e^x$ equals its height there. At $x = 1$ the height is $e \approx 2.718282$, and central differences (lesson 2.2.2) settle on the same number.

A picture shows what this means. Every tangent to $e^x$ meets the $x$-axis exactly 1 unit to the left of its point. So the run is 1 and the rise is the height, and the slope is the height divided by 1.

By the constant rule, any multiple $ce^x$ is its own slope too: $\frac{d}{dx}5e^x = 5e^x$. No other exponential $b^x$ is.

Slide the point along $e^x$ below. The slope traced underneath is the same curve again.

Why $e$ is the natural base

Every exponential $b^x$ has a slope proportional to its height. The factor is its slope at 0, which depends on the base.

$e$ is the one base whose slope at 0 is exactly 1, so the factor disappears. That is the sense in which lesson 1.4.2 called $e$ natural. The section on other bases below says where 0.693147 and 1.098612 come from.

$e^{kx}$ and the chain rule

With the chain rule of lesson 2.3.3, the exponent becomes an inner function. Its slope comes out in front, and the exponential stays as it is.

The sign of $k$ decides the shape.

• $k > 0$, growth. $e^{3x}$ has slope $3e^{3x}$.
• $k < 0$, decay. $e^{-2x}$ has slope $-2e^{-2x}$. The slope is negative, and it shrinks as the curve shrinks.
• $k = 0$. $e^{0} = 1$ is a constant, with slope 0.

Divide the slope of $e^{kx}$ by its value and you get $k$, whatever $x$ is. This is the relative rate: the change per unit of $x$ as a fraction of the current size.

So $e^{0.05t}$, with $t$ in years, always grows at 0.05 of its size per year. Case study 2.3.7 uses this for continuous discounting.

The exponent need not be linear. $e^{x^2}$ has slope $2xe^{x^2}$. The bell-shaped $e^{-x^2/2}$ has slope $-xe^{-x^2/2}$, which is about $-0.606531$ at $x = 1$; it returns in lesson 2.6.3.

With the product rule, $xe^x$ has slope $1 \cdot e^x + x \cdot e^x = (1 + x)e^x$. It is 1 at $x = 0$ and 0 at $x = -1$, where the curve is flat for an instant.

$\ln x$ has slope $\frac{1}{x}$

$\ln x$ undoes $e^x$, so its graph is the mirror image of $e^x$ in the line $y = x$ (lessons 1.1.3 and 1.4.3). A mirror in $y = x$ swaps every run with its rise. So it turns a slope $m$ into $\frac{1}{m}$.

On $e^x$, the point at height 2 sits at $x = \ln 2 \approx 0.69$, and its slope is its height, 2. The mirror image is the point $(2, 0.69)$ on $\ln x$, so the slope there is $\frac{1}{2}$.

The same happens at every point. The slope of $\ln x$ at $x$ is one over the height of $e^x$ at the mirror point, and that height is $x$.

Central differences at 2 give 0.500417, 0.500004 and 0.5 for gaps 0.1, 0.01 and 0.001.

Logs of functions

With the chain rule and the outer function $\ln u$, whose slope is $\frac{1}{u}$, the log of a function has a neat derivative.

• $\ln(1 + x^2)$ has slope $\frac{2x}{1 + x^2}$.
• $\ln(5x)$ has slope $\frac{5}{5x} = \frac{1}{x}$, the same as $\ln x$. By the log rule of lesson 1.4.3, $\ln(5x) = \ln 5 + \ln x$, and the constant $\ln 5$ has slope 0.
• $\ln(x^2)$ has slope $\frac{2x}{x^2} = \frac{2}{x}$.
• $(\ln x)^2$ is different: its outer function is $u^2$, so its slope is $2\ln x \cdot \frac{1}{x} = \frac{2\ln x}{x}$.

Because $\frac{g'}{g}$ is a relative rate, a log turns relative change into an ordinary slope. For example, $\ln(e^{kx}) = kx$ has slope $k$, the relative rate of $e^{kx}$.

Negative inputs, and where $\ln$ breaks

$\ln x$ exists only for $x > 0$, so it has no slope at 0 or below. As $x$ shrinks towards 0, the slope $\frac{1}{x}$ grows without limit: 10 at 0.1, 100 at 0.01, 1000 at 0.001. The curve plunges down the $y$-axis.

$\ln|x|$ is defined for negative $x$ as well: it is $\ln x$ mirrored in the $y$-axis. Its slope is also $\frac{1}{x}$, on both sides of 0. At $x = -2$ it is $-0.5$, because there the curve falls.

Lesson 2.5.2 uses this to undo $\frac{1}{x}$.

Other bases

Lesson 1.4.3 gives $b = e^{\ln b}$, so $b^x = e^{x \ln b}$. The chain rule, with inner function $x\ln b$ and inner slope $\ln b$, then gives the slope of $b^x$. Change of base, $\log_b x = \frac{\ln x}{\ln b}$, gives the slope of $\log_b x$.

• $2^x$ has slope $2^x \ln 2 \approx 0.693147 \times 2^x$.
• A base below 1 has a negative log. $\left(\frac{1}{2}\right)^x$ has slope $-\left(\frac{1}{2}\right)^x \ln 2$: it decays.
• $\log_{10} x$ has slope $\frac{1}{x \ln 10}$, which is about 0.434294 at $x = 1$.

Power, exponential or constant?

Three look-alikes take three different rules. Ask where the variable is.

These derivatives, read backwards, give the antiderivatives of lesson 2.5.2. Case study 2.3.7 uses $\frac{d}{dy}e^{-yt} = -te^{-yt}$ to find a bond's duration exactly.

Common mistakes

1. $e^x$ is not a power of $x$. The power rule does not apply: $\frac{d}{dx}e^x \neq xe^{x - 1}$.
2. Keep the chain factor. $e^{-2x}$ gives $-2e^{-2x}$, and $e^{x^2}$ gives $2xe^{x^2}$, not $e^{2x}$.
3. $\ln$ of a sum is not a sum of logs. $\ln(1 + x)$ has slope $\frac{1}{1 + x}$, not $1 + \frac{1}{x}$.
4. A constant inside $\ln$ drops out. $\ln(5x)$ has slope $\frac{1}{x}$, not $\frac{5}{x}$ or $\frac{1}{5x}$.
5. $(\ln x)^2$ is not $\ln(x^2)$. Their slopes are $\frac{2\ln x}{x}$ and $\frac{2}{x}$.
6. Other bases need $\ln b$. $2^x$ gives $2^x \ln 2$, not $x \cdot 2^{x - 1}$.
7. $\ln$ needs a positive input. For negative $x$, use $\ln|x|$, whose slope is also $\frac{1}{x}$.