# Video: Differentiating Polynomial Functions in the Factored Form

Find the first derivative of the function 𝑦 = (𝑥² + 9)(8𝑥 + 3).

02:53

### Video Transcript

Find the first derivative of the function 𝑦 equals 𝑥 squared plus nine multiplied by eight 𝑥 plus three.

Now, there is more than one way that we could approach this question. For example, we could distribute the parentheses to give 𝑦 as a polynomial function of 𝑥, and then apply the power rule of differentiation in order to find its derivative. Instead though, we note that 𝑦 is a product of two polynomials. It’s 𝑥 squared plus nine multiplied by eight 𝑥 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 𝑢 and 𝑣, if 𝑦 is equal to their product, 𝑢𝑣, then the derivative of 𝑦 with respect to 𝑥, d𝑦 by d𝑥, is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. We multiply each of the functions by the derivative of the other and add them together. We can therefore let 𝑢 equal one of our functions, in this case 𝑥 squared plus nine, and 𝑣 equal the other, in this case eight 𝑥 plus three. Although the choice of 𝑢 and 𝑣 is entirely arbitrary as the product rule is symmetrical in 𝑢 and 𝑣.

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 𝑥 and we’re differentiating it with respect to 𝑥, then we multiply by the power and then decrease the power by one. The derivative of 𝑥 squared, then, is two multiplied by 𝑥 to the power of one, or two 𝑥. And the derivative of nine is just zero, because the derivative of any constant is zero. In the same way, the derivative of eight 𝑥 is eight because it’s eight 𝑥 to the power of zero. And, 𝑥 to the power of zero is just one. And the derivative of three, a constant, is zero. So, we have d𝑢 by d𝑥 equals two 𝑥 and d𝑣 by d𝑥 equals eight.

To find d𝑦 by d𝑥, we need to substitute 𝑢, 𝑣, and their derivatives into the formula for the product rule. We have 𝑢, that’s 𝑥 squared plus nine, multiplied by d𝑣 by d𝑥, which is eight, plus 𝑣, that’s eight 𝑥 plus three, multiplied by d𝑢 by d𝑥, which is two 𝑥. We can then distribute each set of parentheses to give eight 𝑥 squared plus 72 plus 16𝑥 squared plus six 𝑥. And finally, group the like terms, that’s the eight 𝑥 squared and the 16𝑥 squared, to give 24𝑥 squared plus six 𝑥 plus 72.

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