### Video Transcript

Find the derivative of a
vector-valued function π« of π‘ is equal to three minus two π‘π’ plus two π‘ squared
plus three π‘ minus two π£.

The question gives us the
vector-valued function π« of π‘. And it wants us to find the
derivative of this function. First, itβs worth pointing out
thereβs a lot of different notations for vectors. For example, weβve used a half
arrow notation on π« to represent that itβs a function which outputs vectors.

Weβve also used a hat notation on
the vectors π’ and π£ to represent that these are unit vectors. But thereβs a lot of different
other notations you may have seen. For example, we can represent
vectors as bold letters or with a full arrow notation or with an underline. It doesnβt matter how we represent
vectors since these all mean roughly the same thing.

Letβs now go back to the
question. We want to find the derivative of a
vector-valued function. To do this, letβs consider the
vector-valued function π of π‘, which is some function π₯ of π‘π’ plus some
function π¦ of π‘π£. Now to find our derivative of π of
π‘ with respect to π‘, we need to differentiate this component-wise. In other words, we need to find
expressions for the derivative of π₯ of π‘ with respect to π‘ and the derivative of
π¦ of π‘ with respect to π‘.

We get that π prime of π‘ is equal
to π₯ prime of π‘π’ plus π¦ prime of π‘π£. So to find the derivative of π« of
π‘ given to us in the question, we need to find the derivative of three minus two π‘
with respect to π‘ and the derivative of two π‘ squared plus three π‘ minus two with
respect to π‘.

So letβs just evaluate both of
these derivatives. First, three minus two π‘ is a
linear function. So its derivative with respect to
π‘ will be the coefficient of π‘, which in this case is equal to negative two. Next, we need to find the
derivative of two π‘ squared plus three π‘ minus two with respect to π‘. We can do this term by term by
using the power rule for differentiation. We want to multiply by our exponent
of π‘ and then reduce this exponent by one. This gives us four π‘ plus
three.

And now weβre ready to find our
expression for π« prime of π‘. This is the derivative of three
minus two π‘ with respect to π‘π’ plus the derivative of two π‘ squared plus three
π‘ minus two with respect to π‘π£. And of course, weβve already showed
that our first component simplifies to give us negative two and our second component
simplifies to give us four π‘ plus three.

So by using these two expressions,
we were able to show the derivative of the vector-valued function π« of π‘ is equal
to three minus two π‘π’ plus two π‘ squared plus three π‘ minus two π£ with respect
to π‘ is negative two π’ plus four π‘ plus three π£.