Exponentials, Logarithms, and Inverses

Exponential growth describes a quantity whose instantaneous rate is proportional to its current size. Before using that model, we need the derivative of an exponential and a precise account of the inverse relationship between exponentials and logarithms.

We take the natural exponential and logarithm as elementary functions with their usual continuous inverse relationship and the normalization $\lim_{h\to0}(e^h-1)/h=1$. This characterization of the base $e$ is an input here, not a consequence of the differentiation rules already proved. One can instead construct $\log x$ by an integral and derive exponentials after the fundamental theorem; that alternative ordering leads to the same functions.

Exponentials and inverse rates

Using $e^{x+h}=e^xe^h$ in the difference quotient gives $(e^x)'=e^x$. The chain rule gives $(e^{u(x)})'=e^{u(x)}u'(x)$. For a fixed base $b>0$, write $b^x=e^{x\log b}$, obtaining $(b^x)'=b^x\log b$. If $b=1$, the formula correctly returns zero.

If $y=f(x)$ has a differentiable inverse locally, the identity $f(f^{-1}(y))=y$ and the chain rule imply

$$(f^{-1})'(y)=\frac1{f'(f^{-1}(y))},$$

provided the denominator is nonzero. A standard inverse derivative theorem ensures differentiability of the inverse when $f$ is continuously differentiable near the point with nonzero derivative. The requirement is substantive: $x^3$ has an inverse, but at zero its derivative is zero and the inverse cube root lacks a finite derivative there.

Applying the theorem to $e^x$ gives $(\log x)'=1/x$ for $x>0$. For $x\ne0$, $(\log|x|)'=1/x$ on each half-line. A logarithm of a negative argument remains undefined over the reals; the absolute value supplies a different, well-defined function.

Variable arguments and real powers

For a differentiable nonzero $u$, differentiation of $\log|u|$ gives $u'/u$ wherever $u$ does not vanish. Thus $(\log(x^2+1))'=2x/(x^2+1)$ on all real inputs. For $\log(3x-2)$ we additionally require $x>2/3$.

For a real constant $r$ and $x>0$, define $x^r=e^{r\log x}$. Differentiation yields $rx^{r-1}$. Some rational powers extend to negative arguments and require separate domain care; this proof covers the positive domain uniformly.

When both base and exponent vary, an ordinary power rule is insufficient. For $y=x^x$ on $x>0$, take logarithms:

$$\log y=x\log x,\qquad \frac{y'}y=\log x+1, \qquad y'=x^x(\log x+1).$$

The function is positive on the stated domain, so taking the logarithm is legitimate. For $y=(x^2+1)^{\sin x}$, the same method gives

$$y'=(x^2+1)^{\sin x}\left[\cos x\log(x^2+1)+\frac{2x\sin x}{x^2+1}\right].$$

Inverse trigonometric functions

Restrict sine to $[-\pi/2,\pi/2]$ to define $\arcsin$. At interior inputs $|x|<1$, set $y=\arcsin x$. Then $\sin y=x$, so $\cos y\,y'=1$. On that branch cosine is positive, giving

$$(\arcsin x)'=\frac1{\sqrt{1-x^2}}.$$

Likewise $(\arccos x)'=-1/\sqrt{1-x^2}$ for $|x|<1$. For $y=\arctan x$ in $(-\pi/2,\pi/2)$, $\sec^2y\,y'=1$, hence $(\arctan x)'=1/(1+x^2)$ for every real $x$. The inverse branch determines the sign of a square root; it cannot be ignored.

A growth model

For $P(t)=P_0e^{kt}$ with $P_0>0$, $P'=kP$. The parameter $k$ has units inverse time. Doubling time for $k>0$ solves $e^{kt}=2$, giving $\log2/k$. A proportional growth model assumes a constant rate parameter and no limiting resources or changing conditions. Differentiating the formula does not establish those modeling assumptions.

Exercises

1. Differentiate $e^{-3x^2}$ and $2^{5x}$.
2. Differentiate $\log|x^2-4|$ and state the domain.
3. Differentiate $\arctan(2x)$ and $\arcsin(x/2)$ on their derivative domains.
4. Differentiate $x^{\sin x}$ on $x>0$.
5. A positive quantity grows as $P_0e^{0.04t}$ with $t$ in years. Find its doubling time and its instantaneous relative rate.
6. Why can the inverse derivative formula not give a finite derivative for the cube root at zero?

Worked solutions

1. $-6xe^{-3x^2}$ and $5\log2\,2^{5x}$.
2. $2x/(x^2-4)$, with $x\ne\pm2$.
3. $2/(1+4x^2)$ for all real $x$; $1/[2\sqrt{1-x^2/4}]$ for $|x|<2$.
4. $x^{\sin x}[\cos x\log x+(\sin x)/x]$.
5. $\log2/0.04\approx17.329$ years, and $P'/P=0.04$ per year.
6. The original cube function has zero derivative at zero, so the theorem’s nonzero-denominator hypothesis fails. Direct quotients confirm the inverse has unbounded slope.