### Video Transcript

Calculate the derivative of π« of π‘ dot π¬ of π‘ with respect to π‘ for the vector-valued functions π« of π‘ is equal to the sin of π‘ times π’ plus the cos of π‘ times π£ and π¬ of π‘ is equal to the cos of π‘ times π’ plus the sin of π‘ times π£.

The question gives us two vector-valued functions, π« of π‘ and π¬ of π‘. It wants us to find the derivative with respect to π‘ of π« of π‘ dot π¬ of π‘. Remember, since these are vector-valued functions, the dot does not mean multiply; it means the dot product. Before we move any further with this question, itβs also worth pointing out some notations about vectors.

Thereβs a lot of different ways you can see vectors written. For example, you can see them written with bold notation, with underlines, or with full arrows. In this video, weβre going to use half-arrow notation to represent vectors. Weβre also going to use another notation, which is the hat notation. This represents vectors with unit length. Letβs now recall what we mean by the dot product of two vectors.

We recall ππ’ plus ππ£ dot-producted with ππ’ plus ππ£ will give us ππ plus ππ. In other words, we multiply componentwise and then add the results together. So, letβs use this to find an expression for π« of π‘ dot π¬ of π‘. First, we need to multiply the first components in both expressions together. Thatβs the sin of π‘ times the cos of π‘.

Next, we need to add to this the product of the second components of each of our expressions. Thatβs the cos of π‘ times the sin of π‘. So, we get that π« of π‘ dot-producted with π¬ of π‘ is equal to sin of π‘ cos of π‘ plus cos of π‘ sin of π‘. Of course, since multiplication is commutative, we can simplify this by rearranging our second term and then adding these together. And doing this, we get two times the sin of π‘ times the cos of π‘.

Remember, weβre not done though. The question wants us to differentiate this with respect to π‘. And we could do this directly by using the product rule. However, thereβs another way. Recall the double-angle formula for sine tells us the sin of two π‘ is equivalent to two times the sin of π‘ cos of π‘. In other words, we can write π« of π‘ dot π¬ of π‘ as the sin of two π‘. And this expression is much easier to differentiate.

So, we now have the derivative of π« of π‘ dot π¬ of π‘ with respect to π‘ is equal to the derivative of the sin of two π‘ with respect to π‘. And of course, we know for any constant π, the derivative of the sin of ππ‘ is equal to π times the cos of ππ‘. And in our case, the value of the constant π is equal to two. So, we get two times the cos of two π‘.

Therefore, given the vector-valued functions π« of π‘ is equal to the sin of π‘π’ plus the cos of π‘π£ and π¬ of π‘ is equal to the cos of π‘π’ plus the sin of π‘π£. We were able to show the derivative of π« of π‘ dot π¬ of π‘ with respect to π‘ is equal to two times the cos of two π‘.