# Video: Finding the Value of the Derivative of a Function at a Point given the Integration of the Function

Given that ∫ 𝑓(𝑥) d𝑥 = 𝑥³ − 7𝑥² − 𝑥 + 9 + 𝐶, find 𝑓′(1).

02:53

### Video Transcript

Given that the integral of 𝑓 of 𝑥 with respect 𝑥 is equal to 𝑥 cubed minus seven 𝑥 squared minus 𝑥 plus nine plus 𝐶, find 𝑓 prime of one.

In this question, we’ve been told that the integral of some function is equal to some polynomial. To be able to find the first derivative of the function evaluated when 𝑥 is equal to one, that’s 𝑓 prime of one, we’ll begin by using the fundamental theorem of calculus to simply find an expression for the function itself. And of course, this theorem links the concept of differentiating a function with the concept of integrating a function. We can find a function 𝑓 of 𝑥 by differentiating the function 𝑥 cubed minus seven 𝑥 squared minus 𝑥 plus nine plus 𝐶 with respect to 𝑥. And I’m going to rewrite this using Leibniz’s notation to prevent any confusion.

Here, we recall the fact that the derivative of the sum of a number of terms is equal to the sum of the derivative of each of those terms. So, we see that 𝑓 of 𝑥 is equal to the derivative of 𝑥 cubed with respect to 𝑥 plus the derivative of negative seven 𝑥 squared with respect to 𝑥, and so on. We then recall that the derivative of a general polynomial term, 𝑎𝑥 to the power of 𝑛, for real constants 𝑎 and 𝑛 is 𝑛 times 𝑎 times 𝑥 to the power of 𝑛 minus one. In other words, we multiply by the exponent and then reduce that exponent by one. And this tells us that the derivative of 𝑥 cubed is three 𝑥 squared. We then obtain the derivative of negative seven 𝑥 squared to be two times negative seven 𝑥 to the power of one, which is simply negative 14𝑥.

Another way of saying negative 𝑥 is saying negative 𝑥 to the power of one. We multiply by one and then reduce that power by one. So, that gives us one times negative 𝑥 to the power of zero. But of course, 𝑥 to the power of zero is one. So, we see that the derivative of negative 𝑥 is negative one. And we distribute the parentheses because we don’t really need them here. The derivative of a constant is, of course, zero. So, the derivative of nine and the derivative of 𝐶 with respect to 𝑥 is zero. And we found 𝑓 of 𝑥. It’s three 𝑥 squared minus 14𝑥 minus one.

We’re not quite done, though. In this question, we’re looking to find the first derivative of our function and then evaluate it when 𝑥 is equal to one. So, we use a similar process to differentiate 𝑓 of 𝑥 with respect to 𝑥. The derivative of three 𝑥 squared is two times three 𝑥 to the power of one, which is six 𝑥. The derivative of negative 14𝑥 is negative 14. And of course, the derivative of negative one is zero. So, 𝑓 prime of 𝑥 is equal to six 𝑥 minus 14. And we can simply replace 𝑥 with one to find 𝑓 prime of one. That’s six times one minus 14, which is, of course, equal to negative eight. Given the definition of our function 𝑓 of 𝑥, then 𝑓 prime of one is negative eight.