Video: Finding the Derivative of a Product of Functions

If 𝑓(π‘₯) = 2π‘₯Β², 𝑔(π‘₯) = π‘₯, and β„Ž(π‘₯) = 𝑓(π‘₯) β‹… 𝑔(π‘₯), find β„Žβ€²(3).


Video Transcript

If 𝑓 of π‘₯ equals two π‘₯ squared, 𝑔 of π‘₯ equals π‘₯, and β„Ž of π‘₯ equals 𝑓 of π‘₯ times 𝑔 of π‘₯, find β„Ž prime of three.

Now firstly, let’s just be clear on the definition of this function β„Ž of π‘₯. This dot indicates that we are taking the product of the functions 𝑓 of π‘₯ and 𝑔 of π‘₯, not their composition. There are two different approaches that we can take, so let’s consider them both. In the first approach, we can just find the function β„Ž of π‘₯ explicitly by multiplying 𝑓 of π‘₯ and 𝑔 of π‘₯ together. They aren’t particularly complex functions, so this isn’t particularly challenging to do. 𝑓 of π‘₯ is two π‘₯ squared and 𝑔 of π‘₯ is π‘₯. So, the product of these two functions, two π‘₯ squared multiplied by π‘₯, is two π‘₯ cubed. And so, we have an explicit definition for the function β„Ž of π‘₯.

In order to find β„Ž prime of π‘₯, we recall the power rule of differentiation which tells us that the derivative of the general term of the form π‘Žπ‘₯ to the power of 𝑛 is equal to π‘Žπ‘› multiplied by π‘₯ to the power of 𝑛 minus one. We multiply by the original exponent and then reduce the exponent by one. So, β„Ž prime of π‘₯ is equal to two multiplied by three π‘₯ squared, which simplifies to six π‘₯ squared. The final step in this method is to substitute the value of π‘₯ at which we want to evaluate this derivative, that is, π‘₯ equals three. Doing so gives that β„Ž prime of three is equal to six multiplied by three squared. That’s six multiplied by nine, which is equal to 54. And so, by finding an expression for β„Ž of π‘₯ explicitly and then substituting π‘₯ equals three, we found that β„Ž prime of three is equal to 54.

Our second method, which would be particularly useful if the individual functions 𝑓 of π‘₯ and 𝑔 of π‘₯ were quite complicated and we didn’t particularly fancy finding their product, would be to use the product rule of differentiation. This tells us that the derivative with respect to π‘₯ of the product of two differential functions 𝑓 and 𝑔 is equal to 𝑓 multiplied by 𝑔 prime plus 𝑓 prime multiplied by 𝑔. We multiply each function by the derivative of the other and add them together. We have the functions 𝑓 of π‘₯ and 𝑔 of π‘₯; they’re two π‘₯ squared and π‘₯. And by applying the power rule of differentiation, we can find their individual derivatives.

𝑓 prime is equal to four π‘₯, and 𝑔 prime is equal to one. Applying the product rule then, we have that β„Ž prime of π‘₯ is equal to two π‘₯ squared multiplied by one plus four π‘₯ multiplied by π‘₯. That’s two π‘₯ squared plus four π‘₯ squared, which is equal to six π‘₯ squared. And so, we found the same expression for β„Ž prime of π‘₯ using the product rule as we did when we found β„Ž of π‘₯ explicitly. We know then that when we evaluate this derivative π‘₯ equals three, we’ll get the same result. Either of these two methods would be perfectly acceptable. It just depends on the complexity of the two functions 𝑓 of π‘₯ and 𝑔 of π‘₯.

The point that I think this question is really trying to make is that if we were just to multiply together the individual derivatives of 𝑓 of π‘₯ and 𝑔 of π‘₯, so that’s four π‘₯ for 𝑓 prime of π‘₯ and one for 𝑔 prime of π‘₯, then the answer we get, four π‘₯, is not the same as the expression we get for β„Ž prime of π‘₯ when we find its derivative using either of these two methods. We need to be clear then that simply finding the product of the individual derivatives is not the correct way to find the derivative of the product. Using either of the two correct methods then, we found that β„Ž prime of three is equal to 54.

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