# Physics equations/01-Introduction/A:reviewCALCULUS

### Calculus

If f and g are functions of x and a and b are constants, then:   ${\frac {d}{dx}}x^{n}=nx^{n-1}.$ ${\frac {d(af+bg)}{dx}}=a{\frac {df}{dx}}+b{\frac {dg}{dx}}.$ ${\frac {d(fg)}{dx}}={\frac {df}{dx}}g+f{\frac {dg}{dx}}.$ ${\frac {dh}{dx}}={\frac {dh}{dg}}{\frac {dg}{dx}}.$ $\left({\frac {f}{g}}\right)'={\frac {f'g-g'f}{g^{2}}}.$ If y=y(x) and x=x(y) are inverse functions then: ${\frac {dx}{dy}}={\frac {1}{dy/dx}}.$ Indefinite integrals, where $C$ is the arbitrary constant of integration:

$\int \!x^{n}\,dx={\frac {x^{n+1}}{n+1}}+C,\quad (n\neq -1)$ $\int \!x^{-1}\,dx=\ln |x|+C,$ ### Exponential and trigonometric functions

If a is a constant, then: ${\frac {d}{dx}}\left(e^{ax}\right)=ae^{ax}.$ ${\frac {d}{dx}}\left(\ln x\right)={1 \over x},\quad x\neq 0$ $\Rightarrow (\ln f)'={\frac {f'}{f}}\quad$ wherever f is positive.

 $(\sin ax)'=a\cos x\,$ $(\arcsin x)'={1 \over {\sqrt {1-x^{2}}}}\,$ $(\cos ax)'=-a\sin x\,$ $(\arccos x)'=-{1 \over {\sqrt {1-x^{2}}}}\,$ $(\tan x)'=\sec ^{2}x={1 \over \cos ^{2}x}=1+\tan ^{2}x\,$ $(\arctan x)'={1 \over 1+x^{2}}\,$ $(\sec x)'=\sec x\tan x\,$ $(\operatorname {arcsec} x)'={1 \over |x|{\sqrt {x^{2}-1}}}\,$ $(\csc x)'=-\csc x\cot x\,$ $(\operatorname {arccsc} x)'=-{1 \over |x|{\sqrt {x^{2}-1}}}\,$ $(\cot x)'=-\csc ^{2}x={-1 \over \sin ^{2}x}=-(1+\cot ^{2}x)\,$ $(\operatorname {arccot} x)'=-{1 \over 1+x^{2}}\,$ ### Fundamental theorem of calculus

$\int _{a}^{b}{\frac {dF}{ds}}\,ds=\int dF=F|_{a}^{b}=F(b)-F(a)$ $\Rightarrow {\text{If }}\;F(x)=\int _{a}^{x}f(s)\,ds,\,$ ${\text{ then }}\;{\frac {dF}{dx}}=f(x)$ ### Taylor series and Euler's equations

$e^{x}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\dots$ $\sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots$ $\cos x=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+\cdots$ $\Rightarrow \;e^{i\theta }=\cos \theta +i\sin \theta$ 