Video: MATH-DIFF-INT-2018-S1-Q01

If 𝑓′(π‘₯) = π‘₯𝑓(π‘₯) and 𝑓(3) = βˆ’5, find 𝑓″(3).

04:36

Video Transcript

If 𝑓 dash of π‘₯ is equal to π‘₯ times 𝑓 of π‘₯ and 𝑓 of three is equal to negative five, find 𝑓 double dash of three.

To begin, let’s understand the information that the question has given us. Firstly, we’ve been given the general equation for the first derivative of the function 𝑓 of π‘₯. This is denoted as 𝑓 dash of π‘₯. Next, we’ve been given the value of the function 𝑓 of π‘₯ when π‘₯ is equal to three. Finally, the question is asking us to find the value of the second derivative of the function 𝑓 of π‘₯ when π‘₯ is equal to three. And this is denoted as 𝑓 double dash of three.

Now that we understand the terms of our question, our method will be to find an equation for 𝑓 double dash of π‘₯ and then to evaluate this when π‘₯ is equal to three. We can form an equation for 𝑓 double dash of π‘₯ by recalling that this second derivative can be found by differentiating the first derivative with respect to π‘₯. Now, the question has given us an equation for 𝑓 dash of π‘₯. And we can, therefore, use this to substitute in the value into our equation. Doing so, we find that 𝑓 double dash of π‘₯ is equal to d by dπ‘₯ of π‘₯ times our original function, 𝑓 of π‘₯.

If we look at the interior of our brackets, we now see that we must take the derivative of a product, since we have π‘₯ multiplied by 𝑓 of π‘₯. In order to take the derivative of a product, we must use the product rule. When taking the derivative of a product of two different functions of π‘₯, here 𝑔 of π‘₯ and β„Ž of π‘₯, this is equal to the derivative of one of the functions, let’s say 𝑔 of π‘₯, multiplied by the second function, β„Ž of π‘₯, plus the first function, 𝑔 of π‘₯, multiplied by the derivative of the second function, β„Ž of π‘₯.

You may be used to seeing this rule with the letters 𝑓 and 𝑔. But, here, we’ve decided to use β„Ž to avoid confusion with the 𝑓 in our question. Looking back at our working, we now see that we can set the function 𝑔 of π‘₯ to just be π‘₯. And we can set the function β„Ž of π‘₯ to be our original function, 𝑓 of π‘₯. We can now use the product rule to find the derivative that we need. We take the derivative of our first function π‘₯, which is just one, and multiply it by our second function, 𝑓 of π‘₯. We then add this to our first function, π‘₯, multiply it by the derivative of our second function, which is 𝑓 dash of π‘₯. One times 𝑓 of π‘₯ is just 𝑓 of π‘₯. And so, we can get rid of this one in our working.

We can now observe that the derivative 𝑓 dash of π‘₯ has again cropped up in our answer. In order to simplify, we can, therefore, perform the same substitution as earlier, replacing this with π‘₯ times 𝑓 of π‘₯. Doing so, we find our second derivative is equal to 𝑓 of π‘₯ plus π‘₯ times π‘₯ 𝑓 of π‘₯. Let’s now perform some simplifications.

Firstly, π‘₯ times π‘₯ is equal to π‘₯ squared. Now, since both of our terms contain a factor of 𝑓 of π‘₯, we can take this factor out. With this factorization, we now have that 𝑓 double dash of π‘₯ is equal to 𝑓 of π‘₯ times one plus π‘₯ squared. With this equation for 𝑓 double dash of π‘₯, we will now be able to answer the question, which is to find 𝑓 double dash of three. This is the case where π‘₯ is equal to three. Using our equation 𝑓 double dash of three is equal to 𝑓 of three multiplied by one plus three squared, here we have substituted in π‘₯ is equal to three. We can also sub in for 𝑓 of three, since the question has given us that this has a value of negative five.

For our next simplification, we can see that three squared is equal to nine. And the value of our brackets is, therefore, one plus nine, which is equal to 10. The value of 𝑓 double dash of three is, therefore, equal to negative five times 10, which is negative 50. We have now answered the question. And we have found that, for our given function, the value of 𝑓 double dash of three is equal to negative 50.

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