Video: Exploring the Derivatives of Products

Consider the functions 𝑓(π‘₯) = π‘₯, 𝑔(π‘₯) = π‘₯Β². Find 𝑓′(π‘₯) and 𝑔′(π‘₯). Find 𝑓′(π‘₯) 𝑔′(π‘₯). Given that 𝑓(π‘₯)𝑔(π‘₯) = π‘₯Β³, find its derivative.

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Video Transcript

Consider the functions 𝑓 of π‘₯ equals π‘₯, 𝑔 of π‘₯ equals π‘₯ squared. Find 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. Find 𝑓 prime of π‘₯ multiplied by 𝑔 prime of π‘₯. And given that 𝑓 of π‘₯ times 𝑔 of π‘₯ is equal to π‘₯ cubed, find its derivative.

We’ve been given two simple polynomial functions. And we see we’re looking to individually differentiate each function with respect to π‘₯. So, we recall that to differentiate a polynomial function of the term π‘Žπ‘₯ to the 𝑛th power, It’s 𝑛 times π‘Žπ‘₯ to the power of 𝑛 minus one. We multiply the entire term by the exponent and then reduce that exponent by one.

If we say that 𝑓 of π‘₯ is equal to π‘₯ to the power of one, this will help us find its derivative. We multiply the entire term by that exponent. That’s one. And then, we reduce the exponent by one to get zero. So, we have one times π‘₯ to the power of zero. But, of course, we know anything to the power of zero is one. So, we find that 𝑓 prime of π‘₯ is simply equal to one. Similarly, to find 𝑔 prime of π‘₯, we multiply the entire term by two and then reduce the exponent by one to get one. π‘₯ to the power of one is simply π‘₯. So, we find that 𝑔 prime of π‘₯ is equal to two π‘₯.

The second part of this question asks us to find the product of 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. Well, we already saw that 𝑓 prime of π‘₯ is one and 𝑔 prime of π‘₯ is two π‘₯. So, their product is one times two π‘₯, which is, of course, two π‘₯.

Finally, the third part of this question says that given that 𝑓 of π‘₯ times 𝑔 of π‘₯ is π‘₯ cubed, find its derivative. And there are two ways we could go about this. We could use the product rule. And this says that the derivative of the product of two functions is equal to 𝑓 prime of π‘₯ times 𝑔 of π‘₯ plus 𝑓 of π‘₯ times 𝑔 prime of π‘₯. Since we know that 𝑓 of π‘₯ times 𝑔 of π‘₯ is π‘₯ cubed, we find that 𝑓 prime of π‘₯ times 𝑔 of π‘₯ is one times π‘₯ squared and 𝑓 of π‘₯ times 𝑔 prime of π‘₯ is π‘₯ times two π‘₯. That’s π‘₯ squared plus two π‘₯ squared, which is, of course, three π‘₯ squared.

Now, in fact, we didn’t really need to apply the product rule. We could have used the fact that to differentiate a polynomial term, we multiply the entire term by the exponent and then reduce that exponent by one. So, the derivative of π‘₯ cubed is three times π‘₯ squared, or just three π‘₯ squared. What this question is really trying to tell us is that the derivative of 𝑓 of π‘₯ times the derivative of 𝑔 of π‘₯ is not equal to the derivative of 𝑓 times 𝑔 of π‘₯. These are two separate expressions.

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