Welcome to a review of the properties of the derivative of a vector-valued function, part one.
In this video, we'll take a look at proving the property involving the derivative of a dot product.
So here we have the six properties of the derivative of a vector-valued function we're gonna take a look at.
These will look very familiar to some of the properties that we reviewed when we were finding derivatives of basic functions.
This first property just states that if we want to find the derivative of a constant times a vector-valued function R of T,
that'll just equal the constant times the derivative of that vector-valued function.
The second property just states that if we have a sum or difference of vector-valued functions and we want to find the derivative of them,
we just find the derivative of each vector-valued function individually.
This third property states that if we have a function in terms of T times a vector-valued function,
its going to equal the function in terms of T times the derivative of the vector-valued function plus the derivative of the function times the vector-valued function R of T.
This should remind you of the product rule.
But in this case, keep in mind that F of T is a function in terms of T, and R of T is a vector-valued function.
Number four is the property we're gonna prove in this video.
It says the derivative of the dot product of two vectors is equal to
the first vector-valued function dotted with the derivative of the second
plus the derivative of the first vector-valued function dotted with the second.
And then the fifth property states if you want to find the derivative of the cross product of two vector-valued functions
It's gonna be equal to the first vector-valued function crossed with the derivative of the second
plus the derivative of the first vector-valued function crossed with the second vector-valued function.
Again, this looks very similar to the product rule, but now we are talking about cross products.
And this sixth property will remind you of the chain rule when finding derivatives.
Okay, but for this video we're gonna take a look at proving property number four.
In part two of this video, we'll take a look at verifying property number five.
So let's see if we can prove this.
Again, we want to show that the derivative of the dot product of these two vectors, with respect to T,
is equal to the first vector-valued function dotted with the derivative of the second
plus the derivative of the first vector-valued function dotted with the second.
Let's start by defining R of T and U of T.
To save a little bit of work, we'll define these two vectors as only having the X and Y components.
So we'll call this X sub one of T comma Y sub one of T.
And then we'll let
the vector-valued function, U, equal X sub two of T and Y sub two of T.
And to prove this, we are going to have to know the derivatives of these vector-valued functions.
So let's go ahead and list those as well.
The derivative of the vector-valued function R is going to be equal to
X sub one prime of T and Y sub one prime of T.
And so U prime is going to be equal to X sub two prime of T
and Y sub two prime of T.
So we'll start with the left side and see if we can come up with the right side of this property.
So this is going to be equal to the derivative with respect to T of the dot product of R and U.
So we would have X sub one of T times X sub two of T.
Remember this is dot product here.
This is multiplication here
Plus Y sub one of T
times Y sub two of T.
So now to find the derivative of these two products,we'll have to apply the product rule here, and the product rule here as well.
Remeber the product rule says we're gonna take the first function, X sub one of T, multiply it by the derivative of the second, X sub two prime of T
plus the derivative of the first
X sub one prime of T times the second, X sub two of T.
So all of this is just the derivative of this first product.
And now we'll find the derivative of this product.
So we're gonna have plus the same product rule so we'll have Y sub one of T times Y sub two prime of T
plus Y sub one prime of T times Y sub two of T.
Again, this is the derivative of the Y part, applying the product rule.
So believe it or not, the only thing left to do now is to reorder these products meaning we're gonna go ahead and leave this in the first position.
We'll move this product into the second position,
move this into the third position,
move this into the fourth position.
If we take a look at the sum of these first two products,
X sub one of T times X sub two prime of T,
it'd be the X component of R times the X component of U prime.
And then Y one of T times Y two prime of T
would be the Y component of R times the Y component of U prime.
So this
is equal to R of T dotted with U prime of T
If we take a look at the sum of these last two products
X sub one prime of T times X sub two of T
is the X component of R prime times the X component of U.
And Y sub one prime of T times Y sub two of T
is the Y component of R prime times the Y component of U.
So that gives us what we want, this is
R prime of T dotted with U of T.
So there's the proof of property four.
In the next video, we'll take a look at not proving, but verifying property five.  Thank you for watching.
