Derivatives of sin x, cos x, eˣ, and ln x

The derivative of $\sin x$ comes from the definition and two limits from the squeeze theorem lesson, $\lim_{h \to 0} \frac{\sin h}{h} = 1$ and $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$. By the angle-addition formula,

As $h \to 0$ this approaches $\sin x \cdot 0 + \cos x \cdot 1 = \cos x$. The same steps with $\cos(x + h) = \cos x \cos h - \sin x \sin h$ give $-\sin x$ for the derivative of $\cos x$. Both limits need radians, so the rules do too.

For $f(x) = e^x$, the difference quotient factors:

The second factor is the slope of a secant to $y = e^x$ from $(0, 1)$. One way to define $e$ is as the base $b$ for which $\frac{b^h - 1}{h} \to 1$. With $h = 0.01$ that quotient is about $0.70$ for $b = 2$ and $1.10$ for $b = 3$, and about $1.005$ for $b = e \approx 2.718$.

So the tangent to $y = e^x$ at $(0, 1)$ has slope $1$, and the derivative of $e^x$ is $e^x \cdot 1 = e^x$. At every point of the graph, the slope equals the height.

The same factoring works for any base $a > 0$: $\frac{d}{dx}\big[a^x\big] = a^x \cdot \lim_{h \to 0}\frac{a^h - 1}{h}$. Writing $a^h = e^{h \ln a}$ and $k = h\ln a$ (for $a \ne 1$), the limit is $\ln a \cdot \lim_{k \to 0}\frac{e^k - 1}{k} = \ln a$. So $\frac{d}{dx}\big[a^x\big] = a^x \ln a$.

The graph of $y = \ln x$ is the reflection of $y = e^x$ across the line $y = x$, because $\ln x$ undoes $e^x$. Reflecting a line across $y = x$ swaps its rise and run, so its slope becomes the reciprocal. The tangent to $e^x$ at $(\ln a, a)$ has slope $a$, so the tangent to $\ln x$ at $(a, \ln a)$ has slope $\frac{1}{a}$.