# Deriving the rule for direct differentiating Bezier functions

I was reading http://www.idav.ucdavis.edu/education/CAGDNotes/Bernstein-Polynomials/Bernstein-Polynomials.html, the section on the derivative, which gives the derivative for a Bezier curve as:

$B_{k,n}(t) \frac{d}{dt} = n \left ( B_{k-1,n-1}(t) - B_{k,n-1}(t)\right )$

which it then shows the derivation steps for:

$B_{k,n}(t) \frac{d}{dt} = {n \choose k} t^k (1-t)^{n-k} \frac{d}{dt}$

... = $\frac{kn!}{k!(n-k)!} t^{k-1} (1-t)^{n-k} + \frac{(n-k)n!}{k!(n-k)!} t^k (1-t)^{n-1-k}$

... = $\frac{kn!}{k!(n-k)!} t^{k-1} (1-t)^{n-k} + \frac{n(n-1)!}{k!(n-1-k)!} t^k (1-t)^{n-1-k}$

... = $\frac{n(n-1)!}{(k-1)!(n-k)!} t^{k-1} (1-t)^{n-k} + \frac{n(n-1)!}{k!(n-1-k)!} t^k (1-t)^{n-1-k}$

... = $n \left (\frac{(n-1)!}{(k-1)!(n-k)!} t^{k-1} (1-t)^{n-k} + \frac{(n-1)!}{k!(n-1-k)!} t^k (1-t)^{n-1-k} \right )$

... = $n \left ( B_{k-1,n-1}(t) - B_{k,n-1}(t) \right )$

How does this last step go from $(term+term)$ to $(term-term)$? As far as I can tell, the two terms directly map to the lower order Bezier curves, so how does that plus sign turn into a minus?

Doing the steps in between the last and single-to-last, I get this:

... = $n \left (\frac{(n-1)!}{(k-1)!(n-k)!} t^{k-1} (1-t)^{n-k} + \frac{(n-1)!}{k!((n-1)-k)!} t^k (1-t)^{(n-1)-k} \right )$

... = $n \left ( \frac{(n-1)!}{(k-1)!(n-k)!} t^{k-1} (1-t)^{n-k} + B_{k,n-1}(t) \right )$

... = $n \left ( \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!} t^{k-1} (1-t)^{(n-1)-(k-1)} + B_{k,n-1}(t) \right )$

... = $n \left ( B_{k-1,n-1}(t) + B_{k,n-1}(t) \right )$

Since the two terms map directly to the lower order Bezier functions, why does the sign flip?

• The minus sign occurs when $(1-t)$ is differentiated. May 3 '13 at 5:50
• but there is no differentiation in the last step, just a function-for-its-name substitution. Wouldn't the minus sign show up in the very first step if it was due to (1-t) differentiation? May 3 '13 at 13:24

## 1 Answer

The minus sign should be in the second line, due to the chain rule:

$B_{k,n}(t) \frac{d}{dt} = {n \choose k} t^k (1-t)^{n-k} \frac{d}{dt}$

... = ${n \choose k} \cdot f(t) g(t) \frac{d}{dt}$

... = ${n \choose k} \cdot \left ( f'(t) g(t) + f(t)g'(t) \right )$

Due to the chain rule applying to $g(t)$, its derivative is:

$g'(t) = (n-k) \cdot (1-t)^{(n-k)-1} \cdot \left ( (1-t) \frac{d}{dt} \right )$

$... = (n-k) \cdot (1-t)^{(n-k)-1} \cdot -1$

thus:

$f(g)g'(t) = - (n-k) \cdot t^k \cdot (1-t)^{(n-k)-1}$

and thus:

$B_{k,n}(t) \frac{d}{dt} = {n \choose k} \left ( k \cdot t^{k-1} (1-t)^{n-k} - (n-k) \cdot t^k \cdot (1-t)^{(n-k)-1} \right )$

which means there is a minus sign, rather than a plus sign, for the rest of the derivation.