### Video Transcript

By considering the product rule, find a function π so that π prime of π₯ is equal to π to the power of π₯ over root π₯ plus two π to the π₯ multiplied by root π₯.

Weβve been asked to find an expression for a function π given an expression for its derivative π prime of π₯. We are essentially being asked to perform the reverse process of differentiation. And so we need to find the general antiderivative of the function π prime of π₯. This will be a function π of π₯ plus πΆ, where πΆ is any real constant. Now weβre told that we want to find this antiderivative by considering the product rule. We recall that the product rule for two differentiable functions π’ and π£ is given by π’π£ prime is equal to π’ prime π£ plus π’π£ prime. The derivative of the product π’π£ is the derivative of π’ multiplied by π£ plus π’ multiplied by the derivative of π£.

Now, looking at the function π prime of π₯, we see that it is the sum of two expressions, each of which involves some power of π₯ and an exponential term. We should recall that the derivative of π to the power of π₯ is itself. And so the general antiderivative of π to the power of π₯ is π to the power of π₯ plus πΆ. Comparing the expression weβve been given for π prime of π₯ with the definition of the product rule, it appears as if one term in our product, either π’ or π£, is π to the power of π₯ as it can take the role of an antiderivative in one term and its derivative in the other. So letβs assume for now that π£ is equal to π to the power of π₯.

If this is correct and weβve assigned π£ and π£ prime the right way around, then this would mean that in the first term, π’ prime is equal to one over root π₯. And in the second term, π’ is equal to two root π₯. But does this work? Well, we need to check. Using laws of exponents, one over root π₯ can be written as π₯ to the power of negative one-half. And two root π₯ can be written as two π₯ to the power of one-half. Then recalling the power rule of differentiation, we know that the derivative of π₯ to the power of π plus one over π plus one is π₯ to the power of π, where π is a real constant not equal to negative one. And so it follows that the general antiderivative of π₯ to the power of π, where π is again not equal to negative one, is π₯ to the power of π plus one over π plus one plus the constant of antidifferentiation πΆ.

Finding the general antiderivative of what we suspect π’ prime to be, which is π₯ to the power of negative one-half, we have π₯ to the power of negative one-half plus one over negative one-half plus one plus πΆ. Negative a half plus one is positive a half. And dividing by a half is equivalent to multiplying by two. So if π’ prime is π₯ to the power of negative one-half, then its antiderivative π’ is two π₯ to the power of one-half plus πΆ. And this is indeed equal to what we suspect π’ should be when πΆ is equal to zero.

So by comparing the expression for π prime of π₯ with the product rule, weβve found that this is satisfied if π’ is equal to two root π₯. π’ prime is then equal to one over root π₯, π£ is equal to π to the power of π₯, and π£ prime is also equal to π to the power of π₯. The function π of π₯ is then the product of π’ and π£. Thatβs two root π₯ multiplied by π to the power of π₯. And we must include at this stage a general constant of antidifferentiation πΆ.

So by considering the product rule, we found the function π of π₯ such that π prime of π₯ is equal to π to the power of π₯ over root π₯ plus two π to the power of π₯ multiplied by root π₯ to be π of π₯ equals two root π₯π to the power of π₯ plus πΆ.