# Question Video: Finding the Derivative of a Cubic Function Mathematics • Higher Education

Find the derivative of the function π(π‘) = 2π‘Β³ + 8π‘ β 1 using the definition of the derivative.

03:25

### 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, letβs 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. Letβs 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. Weβll now start simplifying our limit. First, weβll 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, weβll 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 couldβve 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, weβve 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, weβve 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.