# Question Video: Differentiating a Vector-Valued Function Given in Column Vector Notation Mathematics • Higher Education

Find the derivative of the following vector-valued function: 𝐫(𝑡) = [5𝑡² + 3𝑡 and 2𝑒^(6𝑡) and 5 cos (𝑡)].

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### Video Transcript

Find the derivative of the following vector-valued function: 𝐫 of 𝑡 is equal to five 𝑡 squared plus three 𝑡 two 𝑒 to the power of six 𝑡 five cos of 𝑡.

In this question, we’re given a vector-valued function 𝐫 of 𝑡. In fact, we can see we’re given this in column vector notation. And we’re asked to find the derivative of this vector-valued function. There’s actually a lot of different ways we could do this. However, all of them end up being more or less the same. All we need to do is differentiate each component of our vector separately. It doesn’t matter if you prefer working with vectors in the form 𝐢, 𝐣, or 𝐤 or with column vectors. We can do either.

In this video, we’re going to use column vector notation. So we need to differentiate each component separately. Let’s start with the first component. That’s five 𝑡 squared plus three 𝑡. So we need to differentiate five 𝑡 squared plus three 𝑡 with respect to 𝑡. We can do this term by term by using the power rule for differentiation. We want to multiply by our exponent of 𝑡 and then reduce this exponent by one. This gives us 10𝑡 plus three. And this will be the first component in our vector-valued function 𝐫 prime of 𝑡. We’ll write this as a column vector because 𝐫 of 𝑡 is also given as a column vector. We now want to do the same with our second component function. So we need to differentiate our second component function. That’s two 𝑒 to the power of six 𝑡 with respect to 𝑡.

And to do this, we can recall one of our derivative results for exponential functions. For any real constants 𝑎 and 𝑏, the derivative of 𝑎 times 𝑒 to the power of 𝑏𝑡 with respect to 𝑡 is equal to 𝑎𝑏 times 𝑒 to the power of 𝑏𝑡. In our case, the value of 𝑎 is two and 𝑏 is six. So differentiating our second component function, we get two times six 𝑒 to the power of six 𝑡. And of course, we can simplify this; two times six is equal to 12. So we’ve shown the derivative of the second component function of 𝐫 of 𝑡 is 12𝑒 to the power of six 𝑡. And this will be the second component function in our vector-valued function 𝐫 prime of 𝑡.

Finally, we just need to do the same for our third component function. So we need to find the derivative of the third component function of 𝐫 of 𝑡. That’s the derivative of five cos of 𝑡 with respect to 𝑡. And we can do this by recalling one of our standard trigonometric derivative results. For any real constant 𝑎, the derivative of 𝑎 cos of 𝑡 with respect to 𝑡 is equal to negative 𝑎 times the sin of 𝑡. So we can differentiate five cos of 𝑡 with respect to 𝑡 by setting our value of 𝑎 equal to five. This gives us negative five times the sin of 𝑡. And then this will be the third and final component of our vector-valued function 𝐫 prime of 𝑡.

And this gives us our final answer. Therefore, by differentiating the vector-valued function 𝐫 of 𝑡 component-wise, we were able to show that 𝐫 prime of 𝑡 will be equal to 10𝑡 plus three 12𝑒 to the power of six 𝑡 negative five sin of 𝑡.