Video: Differentiating Polynomial Functions in the Factored Form

Find the first derivative of the function 𝑦 = (π‘₯Β² + 9)(8π‘₯ + 3).


Video Transcript

Find the first derivative of the function 𝑦 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 𝑦 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 𝑦 is a product of two polynomials. It’s 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 𝑒 and 𝑣, if 𝑦 is equal to their product, 𝑒𝑣, then the derivative of 𝑦 with respect to π‘₯, d𝑦 by dπ‘₯, is equal to 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. We multiply each of the functions by the derivative of the other and add them together. We can therefore let 𝑒 equal one of our functions, in this case π‘₯ squared plus nine, and 𝑣 equal the other, in this case eight π‘₯ plus three. Although the choice of 𝑒 and 𝑣 is entirely arbitrary as the product rule is symmetrical in 𝑒 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 π‘₯ and we’re 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 it’s 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 d𝑒 by dπ‘₯ equals two π‘₯ and d𝑣 by dπ‘₯ equals eight.

To find d𝑦 by dπ‘₯, we need to substitute 𝑒, 𝑣, and their derivatives into the formula for the product rule. We have 𝑒, that’s π‘₯ squared plus nine, multiplied by d𝑣 by dπ‘₯, which is eight, plus 𝑣, that’s eight π‘₯ plus three, multiplied by d𝑒 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, that’s 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 𝑦 is the product of the two differentiable functions, π‘₯ squared plus nine and eight π‘₯ plus three. Then its first derivative, d𝑦 by dπ‘₯, is equal to 24π‘₯ squared plus six π‘₯ plus 72.

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