### Video Transcript

Find the derivative of a vector-valued function π« of π‘ is equal to π‘π’ plus π£ plus π€.

In this question, weβre given a vector-valued function π« of π‘. And weβre asked to determine the derivative of this vector-valued function. The first thing we notice is this is a function in π‘, so we need to differentiate this with respect to π‘. And to do this, we need to recall how we differentiate vector-valued functions. To do this, we just need to differentiate this component-wise. We just differentiate each of our component functions with respect to π‘.

And to highlight this, letβs we rewrite our vector-valued function π« of π‘ in the following way: π‘π’ plus one π£ plus one π€. Of course, we can do this because we know that π£ is the same as one π£ and π€ is the same as one π€. Now that weβve written π« of π‘ in the following form, we can start differentiating our component function. So, letβs start with our first component function. Thatβs π‘. So we want to differentiate π‘ with respect to π‘. Thereβs a few different ways of doing this.

We could write π‘ as π‘ to the first power and then apply the power rule for differentiation. And this would work. However, we could also just remember the derivative of any linear function is the coefficient of π‘. So in this case, thatβs one. Now that weβve found the derivative of our first component function, we know that π« prime of π‘ will start with one π’.

Letβs now move on to our second component function. We need to differentiate one. So we need to find the derivative of one with respect to π‘. Of course, one is a constant, so itβs not varying as the value of π‘ changes. So its derivative evaluates to give us zero. We can then write this into our expression for π« prime of π‘. We now have zero π£.

Finally, we want to do the same for our third component function. We see that thatβs also one. But weβve already found the derivative of one with respect to π‘. Itβs equal to zero. So in our expression for π« prime of π‘, we will also get the term zero π€. Now, we could leave our answer like this. However, we can also simplify.

Zero π£ and zero π€ donβt change the value of π« of π‘. So we can remove these terms. Next, one π’ will just be the same as the vector π’. So our expression for π« prime of π‘ simplifies to just give us the vector π’. Therefore, we were able to show if the vector-valued function π« of π‘ is equal to π‘π’ plus π£ plus π€, then π« prime of π‘ would just be equal to the vector π’.