Skip to content
Geometry and Algebra Math Analysis 1

Calculus formula sheet (Mathematical Analysis 1)

All the formulas of first-year calculus on one page, in syllabus order: algebra, complex numbers, trigonometry, standard limits, derivatives, Taylor expansions, integrals and series.

Updated 19 October 2026 4 min read 0 comments

All the formulas of first-year calculus (Mathematical Analysis 1) on one page, in the order of the course syllabus. It is meant for revision before an exam: to understand where a formula comes from, each section corresponds to a complete lesson listed in the subject’s syllabus.

Algebra, powers and logarithms

(a \pm b)^2 = a^2 \pm 2ab + b^2a^2 - b^2 = (a-b)(a+b)
a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
a^x a^y = a^{x+y}, (a^x)^y = a^{xy}a^{-x} = \frac{1}{a^x}, a^{1/n} = \sqrt[n]{a}
\log_a(xy) = \log_a x + \log_a y\log_a x^k = k\log_a x
\log_a x = \frac{\ln x}{\ln a}a^x = e^{x\ln a}
|x + y| \le |x| + |y||x - c| < r \iff c - r < x < c + r

Complex numbers

z = a + ib = \rho e^{i\theta}\rho = \sqrt{a^2 + b^2}, \bar z = a - ib, z\bar z = \rho^2
e^{i\theta} = \cos\theta + i\sin\thetaz^n = \rho^n e^{in\theta}
\sqrt[n]{z} = \sqrt[n]{\rho}\, e^{i\frac{\theta + 2k\pi}{n}}, k = 0..n-1\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}

Trigonometry

x0\pi/6\pi/4\pi/3\pi/2
\sin x0\frac12\frac{\sqrt2}{2}\frac{\sqrt3}{2}1
\cos x1\frac{\sqrt3}{2}\frac{\sqrt2}{2}\frac120
\tan x0\frac{\sqrt3}{3}1\sqrt3undefined
\sin^2 x + \cos^2 x = 11 + \tan^2 x = \frac{1}{\cos^2 x}
\sin(a \pm b) = \sin a\cos b \pm \cos a\sin b\cos(a \pm b) = \cos a\cos b \mp \sin a\sin b
\sin 2x = 2\sin x\cos x\cos 2x = \cos^2 x - \sin^2 x
\sin^2 x = \frac{1 - \cos 2x}{2}\cos^2 x = \frac{1 + \cos 2x}{2}
\sin(\pi - x) = \sin x\cos(\pi - x) = -\cos x

Standard limits (x → 0)

\frac{\sin x}{x} \to 1\frac{1 - \cos x}{x^2} \to \frac12\frac{\tan x}{x} \to 1
\frac{\ln(1+x)}{x} \to 1\frac{e^x - 1}{x} \to 1\frac{a^x - 1}{x} \to \ln a
\frac{(1+x)^\alpha - 1}{x} \to \alpha(1+x)^{1/x} \to e\left(1 + \frac{a}{x}\right)^x \to e^a (x\to\infty)

Hierarchy at infinity: \ln x \ll x^\alpha \ll a^x \ll x! for \alpha > 0, a > 1.

Derivatives

(x^\alpha)' = \alpha x^{\alpha-1}(e^x)' = e^x(a^x)' = a^x\ln a
(\ln x)' = \frac1x(\sin x)' = \cos x(\cos x)' = -\sin x
(\tan x)' = \frac{1}{\cos^2 x}(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}(\arctan x)' = \frac{1}{1+x^2}

Rules: (fg)' = f'g + fg', \left(\frac fg\right)' = \frac{f'g - fg'}{g^2}, (f(g(x)))' = f'(g(x))\,g'(x), (f^{-1})'(y) = \frac{1}{f'(x)}. Tangent: y = f(x_0) + f'(x_0)(x - x_0).

Mean value theorem: f(b) - f(a) = f'(c)(b - a). L’Hôpital (forms \frac00, \frac\infty\infty): \lim\frac fg = \lim\frac{f'}{g'}.

Maclaurin expansions

e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + o(x^3)\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + o(x^3)
\sin x = x - \frac{x^3}{6} + o(x^4)\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^5)
(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2}x^2 + o(x^2)\frac{1}{1-x} = 1 + x + x^2 + o(x^2)
\arctan x = x - \frac{x^3}{3} + o(x^4)\tan x = x + \frac{x^3}{3} + o(x^4)

Integrals

\int x^\alpha dx = \frac{x^{\alpha+1}}{\alpha+1} + C\int\frac{dx}{x} = \ln|x| + C\int e^x dx = e^x + C
\int\sin x\,dx = -\cos x + C\int\cos x\,dx = \sin x + C\int\frac{dx}{1+x^2} = \arctan x + C
\int\frac{dx}{\sqrt{1-x^2}} = \arcsin x + C\int\frac{f'}{f}dx = \ln|f| + C\int\frac{dx}{\cos^2 x} = \tan x + C

By parts: \int f'g\,dx = fg - \int fg'\,dx. Fundamental theorem: \int_a^b f(x)\,dx = F(b) - F(a), \frac{d}{dx}\int_a^x f(t)\,dt = f(x). Mean value: \frac{1}{b-a}\int_a^b f.

Improper integrals: \int_1^{\infty}\frac{dx}{x^\alpha} converges for \alpha > 1; \int_0^{1}\frac{dx}{x^\alpha} converges for \alpha < 1.

Series

\sum_{n=0}^\infty r^n = \frac{1}{1-r} for |r| < 1\sum\frac{1}{n^\alpha} converges \iff \alpha > 1
Ratio: \frac{a_{n+1}}{a_n} \to \ell < 1 convergesRoot: \sqrt[n]{a_n} \to \ell < 1 converges
Leibniz: b_n \downarrow 0 \Rightarrow \sum(-1)^n b_n convergesNecessary condition: a_n \to 0

Related articles

Subscribe
Notify of
guest
0 Comments
Oldest
Newest Most Voted
0
Would love your thoughts, please comment.x
()
x