### Video Transcript

Find the derivative of the function π of π‘ is equal to two π‘ cubed plus eight π‘ minus one using the definition of the derivative.

The question wants us to find the derivative of the function π of π‘. It wants us to do this by using the definition of the derivative. So, letβs start by recalling the definition of the derivative. We say for some function π of π₯ and some point π₯ naught, the derivative of π of π₯ at π₯ naught is defined by. π prime of π₯ naught is equal to the limit as β approaches zero of π evaluated at π₯ naught plus β minus π evaluated at π₯ naught all divided by β if this limit exists.

And this definition gives us the derivative of our function at a single point. We can then use this to find the derivative of the entire function. Letβs start by finding the derivative of our function π of π‘ at π‘ naught. From our definition of the derivative, we have π prime of π‘ naught is equal to the limit as β approaches zero of π evaluated at π‘ naught plus β minus π evaluated at π‘ naught divided by β.

Next, we can use our definition of the function π of π‘ to rewrite our limit in the following form. Weβll now start simplifying our limit. First, weβll distribute the cube over our set of parentheses. This gives us two π‘ naught cubed plus six π‘ naught squared β plus six π‘ naught β squared plus two β cubed.

Next, weβll multiply out the second pair of parentheses. This gives us eight π‘ naught plus eight β. Finally, we subtract one and multiply our last set of parentheses through by negative one. And, of course, we divide through by β. This gives us the following expression.

We can simplify this expression slightly by canceling some terms. We see we have a plus eight π‘ naught and a minus eight π‘ naught. We also subtract one and add one. Finally, we have two π‘ naught cubed minus two π‘ naught cubed. So, canceling these terms, our limit becomes the limit as β approaches zero of six π‘ naught squared β plus six π‘ naught β squared plus two β cubed plus eight β divided by β.

And now, we can see that every term in our numerator shares a factor of β. And since our limit is as β is approaching zero, β is not equal to zero. So, we can cancel this shared factor of β. And this gives us the limit as β approaches zero of six π‘ zero squared plus six π‘ zero β plus two β squared plus eight. And remember, π‘ zero is just a constant. So, this is a polynomial. And we can evaluate this by direct substitution.

Substituting β is equal to zero, we get six π‘ zero squared plus six π‘ zero times zero plus two times zero squared plus eight. And, of course, we can simplify this to give us six π‘ zero squared plus eight. However, remember, this is only the derivative of our function at the point π‘ zero. And we can notice in our calculations of the derivative, at no point did we use any properties of the constant π‘ zero. The value of π‘ zero couldβve been any real number and our steps would have been exactly the same. And we still would have got six π‘ zero squared plus eight.

In other words, weβve shown for any real value of π‘, the derivative π prime of π‘ is equal to six π‘ squared plus eight. Therefore, by using the definition of the derivative, weβve shown that the function π of π‘ is equal to two π‘ cubed plus eight π‘ minus one has the derivative π prime of π‘ is equal to six π‘ squared plus eight.