Why is this true?
Differentiation of xn
f(x) = xn
f'(x) = nx(n-1)
Can someone explain? Thanks.
Use the first principles.
ddxxn=limh→0(x+h)n−xnh=limh→0xn+(n1)xn−1h+...−xnhother terms are negligible as they evaluate to 0 after taking limit.=limh→0nhxn−1h+lots of negligible terms which evaluates to 0=limh→0nxn−1=nxn−1 YAY
Simple rules of differentiation
Why is this true?
Differentiation of xn
f(x) = xn
f'(x) = nx(n-1)
difference quotient:
ΔyΔx=(x+h)n−xnhΔyΔx=(n0)xn+(n1)xn−1h+(n2)xn−2h2…+(nn)hn−xnh|(n0)=1ΔyΔx=xn+(n1)xn−1h+(n2)xn−2h2…+(nn)hn−xnhΔyΔx=(n1)xn−1h+(n2)xn−2h2…+(nn)hnhΔyΔx=(n1)xn−1+(n2)xn−2h1…+(nn)hn−1|(n1)=nΔyΔx=nxn−1+(n2)xn−2h…+(nn)hn−1
differential quotient:
δyδx=f′(x)=limh→0(nxn−1+(n2)xn−2h…+(nn)hn−1)f′(x)=nxn−1+(n2)xn−2×0…+(nn)×0n−1f′(x)=nxn−1+0…+0f′(x)=nxn−1