Differentiation of Elementary Functions

Every elementary function is differentiable wherever it is defined, and its derivative is again elementary. This checkpoint collects the proofs: starting from two fundamental limits, you can derive the derivatives of the exponential, logarithm, all trigonometric and inverse trigonometric functions, and all hyperbolic and inverse hyperbolic functions.

Two foundational limits

All the proofs below rest on two limits.

Lemma 1. $\displaystyle\lim_{h \to 0} \frac{e^h - 1}{h} = 1$.

Proof. Substitute $u = e^h - 1$, so $h = \ln(1+u)$ and $u \to 0$ as $h \to 0$:

$$\frac{e^h - 1}{h} \;=\; \frac{u}{\ln(1+u)}.$$

From the inequality $\dfrac{x}{1+x} \leq \ln(1+x) \leq x$ (valid for $x > -1$), dividing by $u > 0$:

$$\frac{1}{1+u} \;\leq\; \frac{\ln(1+u)}{u} \;\leq\; 1.$$

As $u \to 0^+$, the left side $\to 1$, so by the squeeze theorem $\ln(1+u)/u \to 1$. The same holds for $u \to 0^-$ by symmetry. Therefore $\dfrac{u}{\ln(1+u)} \to 1$. $\square$

Lemma 2. $\displaystyle\lim_{h \to 0} \frac{\sin h}{h} = 1$.

Proof. For $0 < h < \pi/2$, compare areas in a unit circle:

$$\tfrac{1}{2}\sin h \;\leq\; \tfrac{h}{2} \;\leq\; \tfrac{1}{2}\tan h.$$

Dividing by $\tfrac{1}{2}\sin h > 0$ and inverting:

$$\cos h \;\leq\; \frac{\sin h}{h} \;\leq\; 1.$$

As $h \to 0^+$, $\cos h \to 1$, so by the squeeze theorem $\sin h / h \to 1$. The case $h \to 0^-$ follows from $\sin(-h)/(-h) = \sin h/h$. $\square$

From Lemma 2 we also derive $\displaystyle\lim_{h \to 0}\frac{\cos h - 1}{h} = 0$, since

$$\frac{\cos h - 1}{h} \;=\; -\frac{\sin(h/2)}{h/2} \cdot \sin(h/2) \;\to\; -1 \cdot 0 \;=\; 0.$$

Derivatives of the exponential and logarithm

Natural exponential

$$\frac{d}{dx}(e^x) \;=\; e^x.$$

Proof. By Lemma 1:

$$\lim_{h\to 0}\frac{e^{x+h}-e^x}{h} \;=\; e^x \lim_{h\to 0}\frac{e^h-1}{h} \;=\; e^x. \;\square$$

General exponential

For $a > 0$, $a \neq 1$, write $a^x = e^{x\ln a}$ and apply the chain rule:

$$\frac{d}{dx}(a^x) \;=\; e^{x\ln a} \cdot \ln a \;=\; a^x \ln a.$$

Natural logarithm

Applying the inverse function rule to $e^x$:

$$\frac{d}{dx}(\ln x) \;=\; \frac{1}{e^{\ln x}} \;=\; \frac{1}{x}, \quad x > 0.$$

General power $x^\alpha$

For $x > 0$ and $\alpha \in \mathbb{R}$, write $x^\alpha = e^{\alpha \ln x}$ and apply the chain rule:

$$\frac{d}{dx}(x^\alpha) \;=\; e^{\alpha\ln x} \cdot \frac{\alpha}{x} \;=\; \alpha x^{\alpha - 1}.$$

This extends the integer power rule to all real exponents.

General logarithm

By the change-of-base formula $\log_a x = \ln x / \ln a$:

$$\frac{d}{dx}(\log_a x) \;=\; \frac{1}{x \ln a}.$$

Derivatives of trigonometric functions

Sine

$$\frac{d}{dx}(\sin x) \;=\; \cos x.$$

Proof. Using the addition formula $\sin(x+h) = \sin x \cos h + \cos x \sin h$:

$$\frac{\sin(x+h)-\sin x}{h} \;=\; \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}.$$

As $h \to 0$: $(\cos h - 1)/h \to 0$ and $(\sin h)/h \to 1$ by the lemmas, so the limit is $\cos x$. $\square$

Cosine

$$\frac{d}{dx}(\cos x) \;=\; -\sin x.$$

Proof. Using $\cos(x+h) = \cos x \cos h - \sin x \sin h$:

$$\frac{\cos(x+h)-\cos x}{h} \;=\; \cos x \cdot \frac{\cos h-1}{h} - \sin x \cdot \frac{\sin h}{h} \;\to\; -\sin x. \;\square$$

Tangent and cotangent

By the quotient rule applied to $\tan x = \sin x / \cos x$ and $\cot x = \cos x / \sin x$:

$$\frac{d}{dx}(\tan x) \;=\; \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \;=\; \sec^2 x, \qquad x \neq \tfrac{\pi}{2}+n\pi,$$ $$\frac{d}{dx}(\cot x) \;=\; -\csc^2 x, \qquad x \neq n\pi.$$

Derivatives of inverse trigonometric functions

All four follow from the inverse function rule and Pythagorean identities.

Function	Domain	Derivative
$\arcsin x$	$(-1,1)$	$\dfrac{1}{\sqrt{1-x^2}}$
$\arccos x$	$(-1,1)$	$\dfrac{-1}{\sqrt{1-x^2}}$
$\arctan x$	$\mathbb{R}$	$\dfrac{1}{1+x^2}$
$\text{arccot}\, x$	$\mathbb{R}$	$\dfrac{-1}{1+x^2}$

The last entry: $(\cot x)' = -\csc^2 x = -(1+\cot^2 x)$, so at $x = \cot t$ the inverse function rule gives $(\text{arccot}\, x)' = -1/(1+x^2)$.

Derivatives of hyperbolic functions

Recall $\sinh x = \dfrac{e^x-e^{-x}}{2}$ and $\cosh x = \dfrac{e^x+e^{-x}}{2}$. Differentiating term by term using $(e^x)' = e^x$ and $(e^{-x})' = -e^{-x}$:

$$\frac{d}{dx}(\sinh x) \;=\; \cosh x, \qquad \frac{d}{dx}(\cosh x) \;=\; \sinh x.$$

By the quotient rule on $\tanh x = \sinh x / \cosh x$:

$$\frac{d}{dx}(\tanh x) \;=\; \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} \;=\; \text{sech}^2 x.$$

Similarly, $(\text{coth}\, x)' = -\text{csch}^2 x$ for $x \neq 0$.

Derivatives of inverse hyperbolic functions

The logarithmic representations of inverse hyperbolic functions make their derivatives easy to compute.

Arsinh. Since $\text{arsinh}\, x = \ln(x+\sqrt{x^2+1})$:

$$(\text{arsinh}\, x)' \;=\; \frac{1+\tfrac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}} \;=\; \frac{\sqrt{x^2+1}+x}{(x+\sqrt{x^2+1})\sqrt{x^2+1}} \;=\; \frac{1}{\sqrt{x^2+1}}.$$

Arcosh. For $x > 1$, $\text{arcosh}\, x = \ln(x+\sqrt{x^2-1})$:

$$(\text{arcosh}\, x)' \;=\; \frac{1+\tfrac{x}{\sqrt{x^2-1}}}{x+\sqrt{x^2-1}} \;=\; \frac{1}{\sqrt{x^2-1}}.$$

Artanh. For $|x| < 1$, $\text{artanh}\, x = \tfrac{1}{2}\ln\tfrac{1+x}{1-x}$:

$$(\text{artanh}\, x)' \;=\; \tfrac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right) \;=\; \frac{1}{1-x^2}.$$

Summary

Function	Derivative	Domain
$e^x$	$e^x$	$\mathbb{R}$
$a^x$	$a^x \ln a$	$\mathbb{R}$
$\ln x$	$1/x$	$x>0$
$\log_a x$	$1/(x\ln a)$	$x>0$
$x^\alpha$	$\alpha x^{\alpha-1}$	$x>0$
$\sin x$	$\cos x$	$\mathbb{R}$
$\cos x$	$-\sin x$	$\mathbb{R}$
$\tan x$	$\sec^2 x$	$x\neq\pi/2+n\pi$
$\cot x$	$-\csc^2 x$	$x\neq n\pi$
$\arcsin x$	$1/\sqrt{1-x^2}$	$(-1,1)$
$\arccos x$	$-1/\sqrt{1-x^2}$	$(-1,1)$
$\arctan x$	$1/(1+x^2)$	$\mathbb{R}$
$\text{arccot}\, x$	$-1/(1+x^2)$	$\mathbb{R}$
$\sinh x$	$\cosh x$	$\mathbb{R}$
$\cosh x$	$\sinh x$	$\mathbb{R}$
$\tanh x$	$\text{sech}^2 x$	$\mathbb{R}$
$\text{coth}\, x$	$-\text{csch}^2 x$	$x\neq 0$
$\text{arsinh}\, x$	$1/\sqrt{x^2+1}$	$\mathbb{R}$
$\text{arcosh}\, x$	$1/\sqrt{x^2-1}$	$x>1$
$\text{artanh}\, x$	$1/(1-x^2)$	$\lvert x\rvert<1$