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# Bernstein Function Derivatives

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I've been reading about bezier curves and the basis, or Bernstein, function which is defined as: Jn,i(t) = (n, i)t^i(1 - t)^n-i. Since I'm not very good at differentiating could some tell me how they got this first derivative: J'n, i(n, i) = i - nt/t(1-t) ? If you can't tell me how they got the first derivative could someone at least tell me what "differentiation technique" was used? Thanks, Bino Edited by - Bino on March 20, 2001 10:14:59 PM

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well i remember right from discrete math, n choose k is actually

n!/k!(n-k)! therefore,

if (n!/k!(n-k)!)*x^k(1-x)^n-k is the nth degree polynomial you wish to find the derivative of, then simply use the power rule and the rule for multiplication(err...forgot the name, duh) and find the derivative directly, remember (n,k) is actually a constant!

eventually the derivative ends up being
n(C(k-1,n-1)(t) - C(k,n-1)(t))

or d/dx =

n((n-1)!/((k-1)!*((n-1)-(k-1))!) - (n-1)!/k!((n-1)-k)!

i think thats right anyhow, ok later

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Wow...math looks even more ugly in pure text.

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Wow...math looks even more ugly in pure text.

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