Calculus II More about e¶ e=(1+dx)1dx=limΔx→0(1+Δx)1Δx=limN→∞(1+1N)N ex=limN→∞(1+1N)Nx=limN→∞(1+xNx)Nx Nx→∞ N→∞ ex=limN→∞(1+xN)N ex=(1+xdx)1dx ∑n=0∞Cnxn=ex ∑n=0∞Cnnxn−1=ex=∑n=0∞Cnxn ∑n=0∞Cnnxn−1=∑n=−1∞Cn+1(n+1)xn=c00x−1+∑n=0∞Cn+1(n+1)xn=∑n=0∞Cnxn More about e to the x¶ Cn+1(n+1)=Cn Cn+1=Cnn+1 Cn=Cn−1n e0=1 C0=1 Cn=1n! ex=∑n=0∞xnn! e=∑n=0∞1n! ∑n=0∞Cnxn=ex ∑n=0∞Cnnxn−1=ex=∑n=0∞Cnxn ∑n=0∞Cnnxn−1=∑n=−1∞Cn+1(n+1)xn=c00x−1+∑n=0∞Cn+1(n+1)xn=∑n=0∞Cnxn Cn+1(n+1)=Cn Cn+1=Cnn+1 Cn=Cn−1n e0=1 C0=1 Cn=1n! ex=∑n=0∞xnn! quoteint rule¶ (fg)′=? f(x)g(x)=h(x) g(x)h(x)=f(x) (g(x)h(x))′=f′(x) g′(x)h(x)+g(x)h′(x)=f′(x) g(x)h′(x)=f′(x)−g′(x)h(x) g(x)h′(x)=f′(x)−g′(x)f(x)g(x) g(x)h′(x)=f′(x)−f(x)g′(x)g(x) g(x)h′(x)=f′(x)g(x)g(x)−f(x)g′(x)g(x) g(x)h′(x)=f′(x)g(x)−f(x)g′(x)g(x) h′(x)=f′(x)g(x)−f(x)g′(x)g2(x) (fg)′=f′g−fg′g2