Início/Cálculo diferencial/Derivadas de ordem superior/Exemplos clássicosSeção 6.2Exemplos clássicosFórmulaCopiar LaTeXdndxn[ekx]=knekx\frac{d^n}{dx^n}[e^{kx}]=k^ne^{kx}dxndn[ekx]=knekxVer detalhe →FórmulaCopiar LaTeXdndxn[sen(kx)]=knsen (kx+nπ2)\frac{d^n}{dx^n}[\operatorname{sen}(kx)]=k^n\operatorname{sen}\!\left(kx+\frac{n\pi}{2}\right)dxndn[sen(kx)]=knsen(kx+2nπ)Ver detalhe →FórmulaCopiar 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)Ver detalhe →FórmulaCopiar 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∈N0Ver detalhe →
FórmulaCopiar LaTeXdndxn[sen(kx)]=knsen (kx+nπ2)\frac{d^n}{dx^n}[\operatorname{sen}(kx)]=k^n\operatorname{sen}\!\left(kx+\frac{n\pi}{2}\right)dxndn[sen(kx)]=knsen(kx+2nπ)Ver detalhe →
FórmulaCopiar 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)Ver detalhe →
FórmulaCopiar 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∈N0Ver detalhe →