Question Video: Using the Product Rule | Nagwa Question Video: Using the Product Rule | Nagwa

# Question Video: Using the Product Rule Mathematics • Second Year of Secondary School

## Join Nagwa Classes

By considering the product rule, find a function π so that πβ²(π₯) = (π^(π₯)/β(π₯)) + 2π^(π₯)β(π₯).

04:07

### 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₯.

Weβve 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 πΆ, where πΆ is any real constant. Now weβre 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 πΆ. Comparing the expression weβve 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 letβs assume for now that π£ is equal to π to the power of π₯.

If this is correct and weβve 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 πΆ.

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 πΆ. 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 πΆ. And this is indeed equal to what we suspect π’ should be when πΆ is equal to zero.

So by comparing the expression for π prime of π₯ with the product rule, weβve 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£. Thatβs two root π₯ multiplied by π to the power of π₯. And we must include at this stage a general constant of antidifferentiation πΆ.

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 πΆ.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions