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

Find the derivative of a vector-valued function π«(π‘) = (3 β 2π‘)π’ + (2π‘Β² + 3π‘ β 2)π£.

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### 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, itβs worth pointing out thereβs a lot of different notations for vectors. For example, weβve used a half arrow notation on π« to represent that itβs a function which outputs vectors.

Weβve also used a hat notation on the vectors π’ and π£ to represent that these are unit vectors. But thereβ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 doesnβt matter how we represent vectors since these all mean roughly the same thing.

Letβs now go back to the question. We want to find the derivative of a vector-valued function. To do this, letβs 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 letβs 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 weβre 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, weβve 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£.