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Calculus Basics

1 min read calculus mathematics derivatives

Limits

The limit of a function $f(x)$ as $x$ approaches $a$ is the value that $f(x)$ gets arbitrarily close to.

$$ \lim_{x \to a} f(x) = L $$

Formal Definition

For every $\varepsilon > 0$, there exists a $\delta > 0$ such that:

$$ 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon $$

Derivatives

The derivative of $f$ at $x$ is defined as the limit:

$$ f’(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Power Rule

$$ \frac{d}{dx} x^n = n x^{n-1} $$

Product Rule

If $u(x)$ and $v(x)$ are differentiable, then:

$$ \frac{d}{dx}[u(x) \cdot v(x)] = u’(x)v(x) + u(x)v’(x) $$

Chain Rule

$$ \frac{d}{dx} f(g(x)) = f’(g(x)) \cdot g’(x) $$

Common Derivatives

$f(x)$ $f’(x)$
$x^n$ $n x^{n-1}$
$\sin x$ $\cos x$
$\cos x$ $-\sin x$
$e^x$ $e^x$
$\ln x$ $\frac{1}{x}$

Integrals

The definite integral of $f$ from $a$ to $b$:

$$ \int_a^b f(x),dx = F(b) - F(a) $$

where $F’(x) = f(x)$.

Fundamental Theorem of Calculus

$$ \frac{d}{dx} \int_a^x f(t),dt = f(x) $$

Practice Problem

Evaluate the limit:

$$ \lim_{x \to 0} \frac{\sin x}{x} $$

Answer: This famous limit equals $1$. It can be proven using the squeeze theorem with the inequality $\cos x \leq \frac{\sin x}{x} \leq 1$ for small $x$.