Home/Calcolo differenziale/Derivate di ordine superiore/Esempi classiciSezione 6.2Esempi classiciFormulaCopia LaTeXdndxn[ekx]=knekx\frac{d^n}{dx^n}[e^{kx}]=k^ne^{kx}dxndn[ekx]=knekxVedi dettaglio →FormulaCopia LaTeXdndxn[sin(kx)]=knsin (kx+nπ2)\frac{d^n}{dx^n}[\sin(kx)]=k^n\sin\!\left(kx+\frac{n\pi}{2}\right)dxndn[sin(kx)]=knsin(kx+2nπ)Vedi dettaglio →FormulaCopia LaTeXdndxn[cos(kx)]=kncos (kx+nπ2)\frac{d^n}{dx^n}[\cos(kx)]=k^n\cos\!\left(kx+\frac{n\pi}{2}\right)dxndn[cos(kx)]=kncos(kx+2nπ)Vedi dettaglio →FormulaCopia LaTeXdndxn[xm]={m(m−1)⋯(m−n+1) xm−n,n≤m0,n>m, m∈N0\frac{d^n}{dx^n}[x^m]=\begin{cases} m(m-1)\cdots(m-n+1)\,x^{m-n}, & n\le m\\ 0, & n>m,\ m\in\mathbb{N}_0 \end{cases}dxndn[xm]={m(m−1)⋯(m−n+1)xm−n,0,n≤mn>m, m∈N0Vedi dettaglio →
FormulaCopia LaTeXdndxn[ekx]=knekx\frac{d^n}{dx^n}[e^{kx}]=k^ne^{kx}dxndn[ekx]=knekxVedi dettaglio →
FormulaCopia LaTeXdndxn[sin(kx)]=knsin (kx+nπ2)\frac{d^n}{dx^n}[\sin(kx)]=k^n\sin\!\left(kx+\frac{n\pi}{2}\right)dxndn[sin(kx)]=knsin(kx+2nπ)Vedi dettaglio →
FormulaCopia LaTeXdndxn[cos(kx)]=kncos (kx+nπ2)\frac{d^n}{dx^n}[\cos(kx)]=k^n\cos\!\left(kx+\frac{n\pi}{2}\right)dxndn[cos(kx)]=kncos(kx+2nπ)Vedi dettaglio →
FormulaCopia LaTeXdndxn[xm]={m(m−1)⋯(m−n+1) xm−n,n≤m0,n>m, m∈N0\frac{d^n}{dx^n}[x^m]=\begin{cases} m(m-1)\cdots(m-n+1)\,x^{m-n}, & n\le m\\ 0, & n>m,\ m\in\mathbb{N}_0 \end{cases}dxndn[xm]={m(m−1)⋯(m−n+1)xm−n,0,n≤mn>m, m∈N0Vedi dettaglio →