I am trying to figure out how to take the derivative of the following quadratic Bezier equation, with respect to 't' for the set of numbers between $0$ and $1$. I understand how to take the derivative of basic equations, but this is proving confusing.
This is the derivative provided by the wiki article on Bezier curves, but I am unable to determine how they created it. A walk-through of the steps involved, or help locating information on solving this equation would be appreciated.
I had originally thought the first term, $(1-t)^2P_0$, was handled like a simple equation, multiplying the exponent by the co-efficient of $(1-t)$, giving me $2(1-t)P_0$. But then I started attempting to figure out how the $P_0$ was handled, being subtracted from the $P_1$ variable. Obviously I have failed at that.
It is done in the "natural" way. When we differentiate $(1-t)^2$ we get $-2(1-t)$ (you forgot about the Chain Rule, the derivative of $1-t$ is $-1$, hence the minus sign).
When we differentiate $2(1-t)t$, that is, $2t-2t^2$, we get $2-4t$. And of course when we differentiate $t^2$ we get $2t$.
Putting things together, we get $$(-2+2t)P_0 +(2-4t)P_1+2tP_2.$$ The next step, to get the form given in the OP, is to split the middle term and get $$(-2+2t)P_0 +(2-2t)P_1-2tP_1+2tP_2.$$ We are almost there. The first two terms give $(2-2t)(P_1-P_0)$, and the next two give $(2t)(P_2-P_1)$.
Remark: The slightly artificial splitting of the middle term is perhaps more natural if we differentiate $2t(1-t)$ using the Product Rule. We get $2(1-t)-2t$.