### Video Transcript

Find the first derivative of the function π¦ equals π₯ squared plus nine multiplied by eight π₯ plus three.

Now, there is more than one way that we could approach this question. For example, we could distribute the parentheses to give π¦ as a polynomial function of π₯, and then apply the power rule of differentiation in order to find its derivative. Instead though, we note that π¦ is a product of two polynomials. Itβs π₯ squared plus nine multiplied by eight π₯ plus three. And therefore, this question is an opportunity for us to use the product rule.

The product rule tells us that for two differentiable functions π’ and π£, if π¦ is equal to their product, π’π£, then the derivative of π¦ with respect to π₯, dπ¦ by dπ₯, is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. We multiply each of the functions by the derivative of the other and add them together. We can therefore let π’ equal one of our functions, in this case π₯ squared plus nine, and π£ equal the other, in this case eight π₯ plus three. Although the choice of π’ and π£ is entirely arbitrary as the product rule is symmetrical in π’ and π£.

We then need to find each of their individual derivatives, which we can do using the power rule of differentiation. This tells us that if we have a power of π₯ and weβre differentiating it with respect to π₯, then we multiply by the power and then decrease the power by one. The derivative of π₯ squared, then, is two multiplied by π₯ to the power of one, or two π₯. And the derivative of nine is just zero, because the derivative of any constant is zero. In the same way, the derivative of eight π₯ is eight because itβs eight π₯ to the power of zero. And, π₯ to the power of zero is just one. And the derivative of three, a constant, is zero. So, we have dπ’ by dπ₯ equals two π₯ and dπ£ by dπ₯ equals eight.

To find dπ¦ by dπ₯, we need to substitute π’, π£, and their derivatives into the formula for the product rule. We have π’, thatβs π₯ squared plus nine, multiplied by dπ£ by dπ₯, which is eight, plus π£, thatβs eight π₯ plus three, multiplied by dπ’ by dπ₯, which is two π₯. We can then distribute each set of parentheses to give eight π₯ squared plus 72 plus 16π₯ squared plus six π₯. And finally, group the like terms, thatβs the eight π₯ squared and the 16π₯ squared, to give 24π₯ squared plus six π₯ plus 72.

By applying the product rule then, we found that if π¦ is the product of the two differentiable functions, π₯ squared plus nine and eight π₯ plus three. Then its first derivative, dπ¦ by dπ₯, is equal to 24π₯ squared plus six π₯ plus 72.