The portal has been deactivated. Please contact your portal admin.

Question Video: Differentiating a Vector-Valued Function Mathematics • Higher Education

Find the derivative of a vector-valued function 𝐫(𝑑) = (3 βˆ’ 2𝑑)𝐒 + (2𝑑² + 3𝑑 βˆ’ 2)𝐣.

02:32

Video Transcript

Find the derivative of a vector-valued function 𝐫 of 𝑑 is equal to three minus two 𝑑𝐒 plus two 𝑑 squared plus three 𝑑 minus two 𝐣.

The question gives us the vector-valued function 𝐫 of 𝑑. And it wants us to find the derivative of this function. First, it’s worth pointing out there’s a lot of different notations for vectors. For example, we’ve used a half arrow notation on 𝐫 to represent that it’s a function which outputs vectors.

We’ve also used a hat notation on the vectors 𝐒 and 𝐣 to represent that these are unit vectors. But there’s 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 doesn’t matter how we represent vectors since these all mean roughly the same thing.

Let’s now go back to the question. We want to find the derivative of a vector-valued function. To do this, let’s consider the vector-valued function 𝐟 of 𝑑, which is some function π‘₯ of 𝑑𝐒 plus some function 𝑦 of 𝑑𝐣. Now to find our derivative of 𝐟 of 𝑑 with respect to 𝑑, we need to differentiate this component-wise. In other words, we need to find expressions for the derivative of π‘₯ of 𝑑 with respect to 𝑑 and the derivative of 𝑦 of 𝑑 with respect to 𝑑.

We get that 𝐟 prime of 𝑑 is equal to π‘₯ prime of 𝑑𝐒 plus 𝑦 prime of 𝑑𝐣. So to find the derivative of 𝐫 of 𝑑 given to us in the question, we need to find the derivative of three minus two 𝑑 with respect to 𝑑 and the derivative of two 𝑑 squared plus three 𝑑 minus two with respect to 𝑑.

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

And now we’re ready to find our expression for 𝐫 prime of 𝑑. This is the derivative of three minus two 𝑑 with respect to 𝑑𝐒 plus the derivative of two 𝑑 squared plus three 𝑑 minus two with respect to 𝑑𝐣. And of course, we’ve already showed that our first component simplifies to give us negative two and our second component simplifies to give us four 𝑑 plus three.

So by using these two expressions, we were able to show the derivative of the vector-valued function 𝐫 of 𝑑 is equal to three minus two 𝑑𝐒 plus two 𝑑 squared plus three 𝑑 minus two 𝐣 with respect to 𝑑 is negative two 𝐒 plus four 𝑑 plus three 𝐣.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.