Video: Evaluating the Definite Integral of a Root Function

Find the derivative of a vector-valued function π«(π‘) = (1 + π‘Β³)π’ + (5π‘Β² + 1)π£ + (π‘Β³ + 2)π€.

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Video Transcript

Find the derivative of a vector-valued function π« of π‘ equals one plus π‘ cubed π’ plus five π‘ squared plus one π£ plus π‘ cubed plus two π€.

Remember, we can find the derivative of a vector-valued function by taking the derivative of each component. That means weβre individually going to differentiate, with respect to π‘, one plus π‘ cubed, five π‘ squared plus one, and π‘ cubed plus two. We then recall that to differentiate polynomial terms, we multiply the entire term by the exponent and then reduce that exponent by one. The derivative of one is zero, and the derivative of π‘ cubed is three π‘ squared. So, differentiating our component for π’ and we get three π‘ squared.

Next, weβre going to differentiate the component for π£. Thatβs 10π‘ plus zero, which is simply 10π‘. Finally, weβre going to differentiate π‘ cubed plus two with respect to π‘. Of course, the derivative of that constant is zero, so we obtain the derivative of π‘ cubed plus two to be three π‘ squared plus zero or just three π‘ squared. So, the derivative π prime of π‘ is three π‘ squared π’ plus 10π‘ π£ plus three π‘ squared π€.