Video: Differentiating Polynomials Using the Product Rule

Find the first derivative of 𝑓(π‘₯) = (π‘₯⁸ + 4)(3π‘₯ √(π‘₯ βˆ’ 7))(3π‘₯ √(π‘₯ +7)) at π‘₯ = βˆ’1.

02:33

Video Transcript

Find the first derivative of 𝑓 of π‘₯ equals π‘₯ to the power of eight plus four times three π‘₯ root π‘₯ minus seven times three π‘₯ root π‘₯ plus seven at π‘₯ equals negative one.

To answer this question, we have two options. We could use the product rule twice or we could recall the definition of the derivative of the product of three functions. The derivative of 𝑓 times 𝑔 times β„Ž is the derivative of 𝑓 times 𝑔 times β„Ž plus 𝑓 times the derivative of 𝑔 times β„Ž plus 𝑓 times 𝑔 times the derivative of β„Ž. Now since our function is actually 𝑓 of π‘₯, I’ve changed the functions in this formula to be 𝑒 𝑣 and 𝑀. So let’s work out what the functions 𝑒, 𝑣, and 𝑀 actually are. We can say that 𝑒 of π‘₯ is equal to π‘₯ to the power of eight plus four. 𝑣 of π‘₯ is equal to three π‘₯ root π‘₯ minus seven. And 𝑀 of π‘₯ is equal to three π‘₯ root π‘₯ plus seven.

We need to differentiate each of these functions with respect to π‘₯ as per the formula for the product rule with three functions. The derivative of 𝑒 is fairly straightforward. It’s just eight π‘₯ to the power of seven. But what about 𝑣 and 𝑀? Well, we could use the product rule. But actually, we can simply rewrite each of these expressions. We know that the square root of π‘₯ is the same as π‘₯ to power of one-half. And the laws of exponents tell us we can simplify this expression by adding the powers.

And when we do, we see that 𝑣 of π‘₯ can be written as three π‘₯ to the power of three over two minus seven and 𝑀 of π‘₯ is three π‘₯ to the power of three over two plus seven. And this means the derivative of the 𝑣 is three over two times three π‘₯ to the power of one-half or nine over two π‘₯ to the power of one-half. And actually, the derivative of 𝑀 is the same.

We now have everything we need to substitute this into our formula for the product rule. Now, at this point, you might be tempted to jump straight into substituting π‘₯ is equal to negative one into the derivative. However, we have some roots here and that might cause issues. Instead, we carefully distribute each set of parentheses and simplify fully. And when we do, we see that the derivative of 𝑓 of π‘₯ is 99π‘₯ to the power of 10 minus 392π‘₯ to the power of seven plus 108π‘₯ to the power of two.

And we can now evaluate this at π‘₯ is equal to negative one. It’s 99 times negative one to the power of 10 minus 392 times negative one to the power of seven plus 108 times negative one squared which is equal to 599.

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