The Exponential Function

Polynomials grow faster as their degree rises — but every polynomial is eventually overtaken by the exponential function $y = b^x$ (base $b > 1$). At $x = 10^3$, the polynomial $x^{100}$ has reached $10^{300}$ while $10^x$ has reached $10^{1000}$ — no contest, and the gap only widens.

Salient features of $y = b^x$ (for $b > 1$):

1. Domain: all of $\mathbb{R}$; Range: the positive reals.
2. The point $(0, 1)$ is on every exponential graph ($b^0 = 1$).
3. Ever increasing, and approaching (never meeting) the $x$-axis for large negative $x$.

Base $10$ gives the common exponential; base $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \cdots \approx 2.718$ gives the natural exponential $e^x$ — the star of calculus.

The Logarithm

The logarithm inverts the exponential: $\log_b a = x$ means $b^x = a$. So $\log_2 8 = 3$ and $\log_{10} 10000 = 4$. As a function, $\log_b : \mathbb{R}^+ \to \mathbb{R}$ — only positive numbers have logarithms. Base $e$ gives the natural logarithm $\log x$ (our default in calculus).

Working rules (all bases): $\log(xy) = \log x + \log y$, $\log\frac{x}{y} = \log x - \log y$, $\log x^n = n\log x$, and the base-change $\log_a p = \frac{\log_b p}{\log_b a}$.

One subtlety worth a mark: $x = e^{\log x}$ holds only for $x > 0$ — the right side is not even defined otherwise.

The derivative pair

$$\boxed{\;\frac{d}{dx}\left(e^x\right) = e^x \qquad\quad \frac{d}{dx}\left(\log x\right) = \frac{1}{x}\;}$$

$e^x$ is the function differentiation cannot change — its own derivative. Combined with the chain rule these two formulas unlock a large family: $\frac{d}{dx}e^{u(x)} = e^{u(x)}u'(x)$ and $\frac{d}{dx}\log(u(x)) = \frac{u'(x)}{u(x)}$.

Solved Examples

Example 1 — Four chain-rule composites at once

Differentiate w.r.t. $x$: (i) $e^{-x}$, (ii) $\sin(\log x)$ for $x > 0$, (iii) $\cos^{-1}(e^x)$, (iv) $e^{\cos x}$.

Step 1 — (i): $\frac{d}{dx}e^{-x} = e^{-x} \cdot (-1) = -e^{-x}$.

Step 2 — (ii): $\frac{d}{dx}\sin(\log x) = \cos(\log x) \cdot \frac{1}{x} = \frac{\cos(\log x)}{x}$.

Step 3 — (iii): $\frac{d}{dx}\cos^{-1}(e^x) = \frac{-1}{\sqrt{1 - e^{2x}}} \cdot e^x = \frac{-e^x}{\sqrt{1 - e^{2x}}}$.

Step 4 — (iv): $\frac{d}{dx}e^{\cos x} = e^{\cos x} \cdot (-\sin x) = -\sin x \, e^{\cos x}$.

Answer: each is one chain-rule link applied to the derivative pair — outer function's derivative at the inner value, times the inner derivative.

Example 2 — Is $x = e^{\log x}$ always true?

Determine for which real $x$ the identity $x = e^{\log x}$ holds.

Step 1 — domain first: $\log x$ exists only for $x > 0$, so the right side is undefined for $x \leq 0$.

Step 2 — for $x > 0$: taking logarithms of $y = e^{\log x}$ gives $\log y = \log x$, so $y = x$.

Answer: the identity holds exactly for positive $x$ — a favourite true/false trap.

Example 3 — Logarithm of a logarithm

Differentiate $\log(\log x)$, $x > 1$, w.r.t. $x$.

Step 1 — outer log: $\frac{1}{\log x}$ at the inner value.

Step 2 — inner log: times $\frac{1}{x}$.

Answer: $\dfrac{1}{x \log x}$ — the condition $x > 1$ keeps $\log x$ positive so the outer logarithm is defined.

Example 4 — Exponential with a power inside

Differentiate $e^{x^3}$ w.r.t. $x$.

Step 1 — chain rule: $e^{x^3}$ times the derivative of $x^3$.

Answer: $3x^2 e^{x^3}$ — the exponential reproduces itself; only the inner derivative multiplies on.

Example 5 — Inverse trig meets exponential

Differentiate $e^{\sin^{-1} x}$ w.r.t. $x$.

Step 1 — chain rule: $e^{\sin^{-1} x}$ times $\frac{d}{dx}\sin^{-1} x$.

Answer: $\dfrac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}}$, for $-1 < x < 1$ — two sections' formulas snapped together by one chain link.

Example 6 — A log-of-cos-of-exp triple

Differentiate $\log(\cos e^x)$ w.r.t. $x$.

Step 1 — outer log: $\frac{1}{\cos e^x}$ at the inner value.

Step 2 — middle cos: times $-\sin e^x$.

Step 3 — inner exp: times $e^x$.

Answer: $\dfrac{-e^x\sin e^x}{\cos e^x} = -e^x\tan(e^x)$ — three layers, three factors, then a trig clean-up.