### Video Transcript

Find the third derivative of the function π¦ equals π₯ to the fifth power plus five π₯ to the fourth power plus three π₯ cubed plus two π₯ squared minus π₯ minus nine.

A third derivative means that we need to differentiate our function π¦ three times, which we can do step by step. We notice that our function π¦ is a polynomial. And so, we recall, first of all, that in order to find the derivative of a general polynomial term ππ₯ to the π for some real constants π and π, this is equal to πππ₯ to the power of π minus one. We multiply by the exponent and then reduce the exponent by one. Letβs go step by step then. We find the first derivative by differentiating once.

Applying the power rule of differentiation and differentiating term by term. We have that the first derivative of our function π¦, dπ¦ by dπ₯, is equal to five π₯ to the fourth power plus five multiplied by four π₯ cubed plus three multiplied by three π₯ squared plus two multiplied by two π₯ minus one. Remember, the derivative of a constant is just zero. So the derivative of that final term, negative nine, is zero. Simplifying each of the coefficients, we have that dπ¦ by dπ₯ is equal to five π₯ to the fourth power plus 20π₯ cubed plus nine π₯ squared plus four π₯ minus one.

To find the second derivative, d two π¦ by dπ₯ squared, we differentiate again. We apply the power rule of differentiation. And once weβve simplified the constants, we have that the second derivative is equal to 20π₯ cubed plus 60π₯ squared plus 18π₯ plus four. To find the third derivative of our function, we differentiate one more time. Applying the power rule and simplifying the constants. We have that the third derivative of π¦ with respect to π₯ is equal to 60π₯ squared plus 120π₯ plus 18.

Notice that the exponent in each term has reduced by three because weβve differentiated three times. And each time, the exponent reduces by one. The first term initially had an exponent of five. Then we differentiated and we had an exponent of four, then an exponent of three, and then an exponent of two. The terms whose exponents were less than three β thatβs two π₯ squared, negative π₯, and negative nine β have differentiated to zero by the time weβve differentiated three times. Our answer to the problem is that the third derivative of the function π¦ is 60π₯ squared plus 120π₯ plus 18.