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Calculus

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Taylor expansion

  • Taylor expansion: f(x)≈f(x0)+f′(x0)1!(x−x0)+f′′(x0)2!(x−x0)2+...f(x) \approx f(x_0) + \frac{f'(x_0)}{1!}(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + ...

Single-variable calculus

Derivatives:

ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1}

ddxax=axln(a)\frac{d}{dx}a^x = a^{x}ln(a)

ddxln(x)=1/x\frac{d}{dx}ln(x) = 1/x

ddxtan(x)=sec2(x)\frac{d}{dx}tan(x)= sec^2(x)

ddxcot(x)=−csc2(x)\frac{d}{dx}cot(x)= -csc^2(x)

ddxsec(x)=sec(x)tan(x)\frac{d}{dx}sec(x)= sec(x)tan(x)

ddxcsc(x)=−csc(x)cot(x)\frac{d}{dx}csc(x)= -csc(x)cot(x)

∫tan=ln∥sec∥\int tan = ln\|sec\|

∫cot=ln∥sin∥\int cot = ln\|sin\|

∫sec=ln∥sec+tan∥\int sec = ln\|sec+tan\|

∫csc=ln∥csc−cot∥\int csc = ln\|csc-cot\|

∫dua2−u2=sin−1(ua)\int \frac{du}{\sqrt{a^2-u^2}} = sin^{-1}(\frac{u}{a})

∫duuu2−a2=1asec−1(ua)\int \frac{du}{u\sqrt{u^2-a^2}} = \frac{1}{a}sec^{-1}(\frac{u}{a})

∫dua2+u2=1atan−1(ua)\int \frac{du}{a^2+u^2} = \frac{1}{a} tan^{-1}(\frac{u}{a})

Continuous: left limit = right limit = value

Differentiable: continuous and no sharp points / asymptotes

L'Hospital's - for indeterminate forms: (f(x)g(x))′=f′(x)g′(x)(\frac{f(x)}{g(x)})' = \frac{f'(x)}{g'(x)}

Integration by parts: ∫udv=uv−∫duv\int{udv}=uv-\int{duv}, LIATE

Expansions:

ex=∑xnn!e^x = \sum{\frac{x^n}{n!}}

sin(x)=∑0∞(−1)nx2n+1(2n+1)!sin(x) = \sum_0^\infty{\frac{(-1)^n x^{2n+1}}{(2n+1)!}}

cos(x)=∑0∞(−1)nx2n(2n)!cos(x) = \sum_0^\infty{\frac{(-1)^n x^{2n}}{(2n)!}}

Geometric Sum: a1st1−rn+11−ra_{1st}\frac{1-r^{n+1}}{1-r}

Multivariable calculus

  • Polar: r,θ\theta,z
  • Spherical: ρ,θ,ϕ\rho,\theta,\phi
  • Clairut's Thm: Conservative function fxy=fyxf_{xy}=f_{yx}
  • Lagrangian - solves minimize f subject to g = c
    • solution will always be tangent to f
    • ∇f=λ∇g\nabla f = \lambda \nabla g - gives us n constraints
    • remember g = c is a constraint too
    • to do this efficiently, define the Lagrangian L(x,λ)=f−λ⋅gL(x, \lambda) = f - \lambda \cdot g
      • taking deriv wrt λ\lambda and setting = 0 enforces g = c
      • taking deriv wrt other variables and setting = 0 enforces other conditions
      • therefore final eq just becomes ∇L=0\nabla L = 0
© 2026 Kiarash Soleimanzadeh