Inverse Functions, Logarithms, and Exponentials

A function takes an input to an output. Sometimes the useful question runs the other way: given the output, which input produced it? For a bacteria culture whose population $N = f (t)$ is recorded hourly, the biologist may instead want the time $t$ required to reach a given level $N$ . That reversed rule is the inverse function $f^{- 1}$ , and it exists only when $f$ never sends two inputs to the same output. The two families where this reversal is used most are the exponentials $b^{x}$ and the logarithms $log_{b} x$ , inverse to each other, and the derivative of an inverse gives both their derivatives from a single rule.

One-to-one functions

An inverse can only be defined when no output is claimed by two different inputs. If $g (2) = g (3) = 4$ , there is no way to recover the input from the output $4$ .

Geometrically this is a statement about horizontal lines. If some horizontal line $y = c$ met the graph at two points $x_{1}, x_{2}$ , then $f (x_{1}) = f (x_{2}) = c$ with $x_{1} \neq = x_{2}$ , breaking the definition.

For example, $f (x) = x^{3}$ is one-to-one: distinct numbers have distinct cubes, and no horizontal line meets the cubic twice. By contrast $g (x) = x^{2}$ fails, since $g (- 1) = g (1) = 1$ . An increasing function is automatically one-to-one, and so is a decreasing one, because a strictly monotonic graph can never double back to repeat a height.

The Horizontal Line Test. A cubic meets every horizontal line exactly once and is one-to-one; a parabola meets some lines twice and is not.

The inverse function

The inverse reverses the assignment: if $f$ sends $x$ to $y$ , then $f^{- 1}$ sends $y$ back to $x$ . Domain and range trade places,

dom f^{- 1} = ran f, ran f^{- 1} = dom f .

The superscript $- 1$ is notation, not an exponent: $f^{- 1} (x)$ does not mean $1/ f (x)$ . Applying a function and then its inverse returns the starting value, which is the content of the cancellation equations:

f^{- 1} (f (x)) = x for x \in A, f (f^{- 1} (x)) = x for x \in B .

To compute a formula for $f^{- 1}$ , solve $y = f (x)$ for $x$ , then interchange the names of the variables so the independent variable is again called $x$ .

For $f (x) = x^{3} + 2$ , solving $y = x^{3} + 2$ gives $x = 3 y - 2$ , and after interchanging, $f^{- 1} (x) = 3 x - 2$ . The rule cube, then add 2 is reversed by subtract 2, then take the cube root.

The graph of $f^{- 1}$ is the reflection of the graph of $f$ across the line $y = x$ . The point $(a, b)$ lies on the graph of $f$ exactly when $(b, a)$ lies on the graph of $f^{- 1}$ , and $(b, a)$ is the mirror image of $(a, b)$ in the diagonal.

A one-to-one function and its inverse are reflections of each other across the line y = x; the point (a,b) on f maps to (b,a) on the inverse.

A function that fails the Horizontal Line Test can often be salvaged by restricting its domain. The full parabola $y = x^{2}$ has no inverse, but $f (x) = x^{2}$ on $[0, 2]$ is one-to-one, with inverse $f^{- 1} (x) = x$ on $[0, 4]$ .

The derivative of an inverse

Reflecting a graph across $y = x$ interchanges the roles of the horizontal and vertical directions, so it interchanges rise and run. The slope of $f^{- 1}$ at a point is therefore the reciprocal of the slope of $f$ at the mirror point.

The reason is implicit differentiation. Write $y = f^{- 1} (x)$ , so $f (y) = x$ . Differentiating $f (y) = x$ with respect to $x$ and using the chain rule,

f^{'} (y) \frac{d y}{d x} = 1 ⟹ \frac{d y}{d x} = \frac{1}{f ^{'} ( y )} = \frac{1}{d x / d y} .

In Leibniz notation the rule reads $\frac{d y}{d x} = \frac{1}{d x / d y}$ , which is the reciprocal-slope statement written symbolically.

Writing g = f^{-1}, reflection across y = x sends the tangent to f at (a,b), of slope m, to the tangent to g at (b,a), of reciprocal slope 1/m.

The formula gives the derivative of an inverse without ever computing the inverse itself.

Exponential functions

An exponential function has the form $f (x) = b^{x}$ with a positive constant base $b$ ; the variable sits in the exponent. This is not the power function $x^{b}$ , where the variable is the base. For a positive integer $n$ , $b^{n}$ is $b$ multiplied by itself $n$ times; the laws of exponents extend the definition to rational $x$ , and a limiting process fills the irrational values so that $b^{x}$ is continuous everywhere.

Every graph $y = b^{x}$ passes through $(0, 1)$ because $b^{0} = 1$ , and larger bases climb more steeply for $x > 0$ . The end behavior follows from the monotonicity:

b > 1 : x \to \infty lim b^{x} = \infty, x \to - \infty lim b^{x} = 0; 0 < b < 1 : x \to \infty lim b^{x} = 0, x \to - \infty lim b^{x} = \infty.

In either case the $x$ -axis is a horizontal asymptote.

The family y = b^x. Every curve passes through (0,1); bases above 1 increase and bases below 1 decrease, each a reflection of another across the vertical axis.

The number e and the derivative of the exponential

Differentiating $b^{x}$ from the definition exposes a constant that depends only on the base:

f^{'} (x) = h \to 0 lim \frac{b ^{x + h} - b ^{x}}{h} = b^{x} h \to 0 lim \frac{b ^{h} - 1}{h} = f^{'} (0) b^{x} .

The rate of change of an exponential is proportional to its own value, with proportionality constant $f^{'} (0)$ , the slope at the $y$ -intercept. Numerically $f^{'} (0) \approx 0.69$ for $b = 2$ and $\approx 1.10$ for $b = 3$ , so some base between $2$ and $3$ gives slope exactly $1$ . That base is the number $e$ .

With $f^{'} (0) = 1$ , the proportionality formula collapses to the defining property of the natural exponential function: it is its own derivative.

The slope of the tangent to $y = e^{x}$ at any point equals the height of the curve there. No other elementary function reproduces itself under differentiation, which is why $e$ rather than $10$ or $2$ is the natural base for calculus.

For y = e^x the slope of the tangent at each point equals the y-coordinate at that point; here the tangent at (1, e) has slope e.

For example, $\frac{d}{d x} e^{tan x} = e^{tan x} sec^{2} x$ by the chain rule, and the product rule gives $\frac{d}{d x} (e^{- 4 x} sin 5 x) = e^{- 4 x} (5 cos 5 x - 4 sin 5 x)$ . The integral counterpart is $\int e^{x} d x = e^{x} + C$ .

Logarithmic functions

Since $b^{x}$ is one-to-one for $b \neq = 1$ , it has an inverse, the logarithm to base $b$ .

The cancellation equations become

log_{b} (b^{x}) = x for all x \in R, b^{log_{b} x} = x for all x > 0.

As the inverse of the exponential, $log_{b}$ has domain $(0, \infty)$ , range $R$ , and a graph obtained by reflecting $y = b^{x}$ across $y = x$ . Every logarithmic graph passes through $(1, 0)$ because $log_{b} 1 = 0$ . The rapid growth of $b^{x}$ reflects into the very slow growth of $log_{b} x$ . The properties of exponents transpose into the properties of logarithms.

The vertical asymptote of the logarithm is the reflected image of the exponential's horizontal asymptote:

x \to \infty lim log_{b} x = \infty, x \to 0^{+} lim log_{b} x = - \infty (b > 1) .

The natural logarithm

The most convenient base is $e$ ; its logarithm gets the special notation $ln x = log_{e} x$ , the natural logarithm. The defining relations and cancellation equations specialize to

ln x = y \leftrightarrow e^{y} = x, ln (e^{x}) = x, e^{ln x} = x, ln e = 1.

These two cancellation identities solve equations with an unknown in the exponent.

Any logarithm can be rewritten in terms of $ln$ , which is why calculators need only one logarithm key.

The proof is one line: if $y = log_{b} x$ then $b^{y} = x$ ; taking $ln$ gives $y ln b = ln x$ , so $y = ln x / ln b$ . As an application, $log_{8} 5 = ln 5/ ln 8 \approx 0.773976$ .

The natural exponential and natural logarithm are inverses, so their graphs mirror across y = x. Because e^x has slope 1 at (0,1), ln x has slope 1 at (1,0).

Derivatives of logarithms

The natural logarithm is differentiable because it inverts the differentiable function $e^{x}$ . Writing $y = ln x$ , so $e^{y} = x$ , and differentiating implicitly gives $e^{y} d y / d x = 1$ , hence $d y / d x = 1/ e^{y} = 1/ x$ .

The chain-rule form is the one used in practice. For $y = ln (x^{3} + 1)$ ,

\frac{d y}{d x} = \frac{1}{x ^{3} + 1} \cdot 3 x^{2} = \frac{3 x ^{2}}{x ^{3} + 1},

and $\frac{d}{d x} ln (sin x) = \frac{cos x}{sin x} = cot x$ . Extending to absolute values gives a formula valid on both sides of the origin,

\frac{d}{d x} ln ∣ x ∣ = \frac{1}{x}, whose integral form \int \frac{1}{x} d x = ln ∣ x ∣ + C

fills the one gap in the power rule for antiderivatives: $\int x^{n} d x$ gives $x^{n + 1} / (n + 1)$ except when $n = - 1$ , and that missing case is supplied by the logarithm. As one consequence, $\int tan x d x = ln ∣ sec x ∣ + C$ , found by substituting $u = cos x$ .

The general and base- $b$ formulas follow from $ln$ through change of base and the identity $e^{ln b} = b$ .

Function	Derivative	Source
$e^{x}$	$e^{x}$	defining property
$b^{x}$	$b^{x} ln b$	$b^{x} = e^{(ln b) x}$
$ln x$	$1/ x$	inverse of $e^{x}$
$log_{b} x$	$1/ (x ln b)$	change of base

The base- $b$ exponential derivative comes from rewriting $b^{x} = e^{(ln b) x}$ and applying the chain rule:

\frac{d}{d x} b^{x} = \frac{d}{d x} e^{(ln b) x} = e^{(ln b) x} \cdot ln b = b^{x} ln b .

The base- $b$ logarithm derivative is $\frac{d}{d x} log_{b} x = \frac{1}{x ln b}$ , smallest and simplest when $b = e$ so that $ln b = 1$ . That simplicity is the reason calculus prefers natural logarithms.

Logarithmic differentiation

Products, quotients, and powers stacked together are painful to differentiate directly, but taking a logarithm first turns them into sums and differences. The technique is logarithmic differentiation.

The same device proves the power rule for an arbitrary real exponent. Let $y = x^{n}$ with $x \neq = 0$ ; then $ln ∣ y ∣ = n ln ∣ x ∣$ , so $y^{'} / y = n / x$ and $y^{'} = n y / x = n x^{n - 1}$ . Functions of the form $[g (x)]^{h (x)}$ , where both base and exponent vary, need the same treatment or the equivalent rewrite $e^{h (x) ln g (x)}$ ; neither the power rule nor the exponential rule applies on its own.

The number e as a limit

The derivative $ln^{'} (1) = 1$ can be unwound into a limit formula for $e$ itself. From the definition of the derivative at $x = 1$ ,

1 = ln^{'} (1) = x \to 0 lim \frac{ln ( 1 + x ) - ln 1}{x} = x \to 0 lim ln (1 + x)^{1/ x} .

By continuity of the exponential, exponentiating both sides gives

e = x \to 0 lim (1 + x)^{1/ x}, equivalently e = n \to \infty lim (1 + \frac{1}{n})^{n} .

The second form, with $n = 1/ x$ , is the expression that reappears in continuously compounded growth: letting the compounding interval shrink to zero drives the accumulation factor to $e$ .