Video: Using the Product Rule

Suppose that 𝑓(2) = 3, 𝑔(2) = 5, 𝑓′(2) = βˆ’1, and 𝑔′(2) = 6. Evaluate (𝑓(π‘₯)𝑔(π‘₯))β€² βˆ’ 𝑓′(π‘₯)𝑔′(π‘₯) at π‘₯ = 2.

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Video Transcript

Suppose that 𝑓 of two equals three, 𝑔 of two equals five, 𝑓 prime of two equals negative one, and 𝑔 prime of two equals six. Evaluate 𝑓 of π‘₯ 𝑔 of π‘₯ prime, whereas 𝑓 prime of π‘₯ 𝑔 prime of π‘₯ at π‘₯ equals two.

We’ve been given the values of two functions 𝑓 of π‘₯ and 𝑔 of π‘₯ and their derivatives at π‘₯ equals two. And we’re asked to evaluate this more complicated expression involving derivatives. We are able to substitute some values straightaway. We’re told in the question that 𝑓 prime of two is equal to negative one. So when evaluating this expression at π‘₯ equals two, we’ll be able to swap 𝑓 prime of π‘₯ for negative one. We’re also told that 𝑔 prime of two is equal to six. So when evaluating this expression π‘₯ equals two, we’ll be able to swap 𝑔 prime of π‘₯ for six.

What we don’t know is the value of 𝑓 of π‘₯ 𝑔 of π‘₯ all prime when π‘₯ is equal to two. Now, this means we find the product of the functions 𝑓 of π‘₯ and 𝑔 of π‘₯ and then differentiate it. So we need to recall how to find the derivative of a product of functions. We need to use the product rule, which tells us that for two differentiable functions 𝑓 of π‘₯ and 𝑔 of π‘₯ the derivative of their product 𝑓 of π‘₯ 𝑔 of π‘₯ is equal to 𝑓 times 𝑔 prime plus 𝑔 times 𝑓 prime. We multiply each function by the derivative of the other and add these together.

So we now know how to find 𝑓 of π‘₯ 𝑔 of π‘₯ all prime in terms of 𝑓 of π‘₯, 𝑔 of π‘₯, and their derivatives. To evaluate this derivative at π‘₯ equals two, we just need to substitute two for π‘₯. So we have that 𝑓 of two 𝑔 of two all prime is equal to 𝑓 of two multiplied by 𝑔 prime of two plus 𝑔 of two multiplied by 𝑓 prime of two. And each of these values is given in the question. Remember, 𝑓 prime of two is equal to negative one and 𝑔 prime of two is equal to six. We’re also given that 𝑓 of two is equal to three and 𝑔 of two is equal to five.

So we have all of the values we need to evaluate this. So that full expression 𝑓 of π‘₯ 𝑔 of π‘₯ prime minus 𝑓 prime of π‘₯ 𝑔 prime of π‘₯ all evaluated at π‘₯ equals two is equal to three times six plus five times negative one minus negative one times six. That’s 18 minus five minus negative six, which is equal to 19. So by recalling the product rule, which tells us how to find the derivative of a product of two differentiable functions and then substituting the relevant values of 𝑓 of two, 𝑔 of two, 𝑓 prime of two, and 𝑔 prime of two. We found that the value of this expression when π‘₯ is two is 19.

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