00:01
So let's say you want to take the derivative of some vector.
00:04
And for this example, we'll use the vector 5 squared 2t, comma, minus ln2t, and then the final term is minus 5e to minus 4t.
00:19
Well, the way you take a derivative of vectors, you just take the derivative of each of the components of the vector, right? so the derivative of the first term is just 5 times the derivative of.
00:30
The square of 2t.
00:32
Now for this one, i like to think of the square roots as raised to the power of one half, right? so the derivative of this is just one half, two t to power of minus one half, right? multiply the exponent to the front and then you subtract the exponent times the derivative of the inside, right? because this is a chain rule.
00:58
So then these two cancel out.
01:02
And you get 2t, then minus one half is just one over.
01:05
The square of 2t, right? and then derivative of the second term.
01:14
So this is gonna be another chain rule.
01:17
Derivative of ln is just one over the function.
01:20
So minus one over two t.
01:22
Since this is a chain rule, we have to multiply by the derivative of whatever's inside the function...