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

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By considering the product rule, find a function 𝑓 so that 𝑓′(π‘₯) = (𝑒^(π‘₯)/√(π‘₯)) + 2𝑒^(π‘₯)√(π‘₯).

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Video Transcript

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

We’ve been asked to find an expression for a function 𝑓 given an expression for its derivative 𝑓 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 𝑓 prime of π‘₯. This will be a function 𝑓 of π‘₯ plus 𝐢, where 𝐢 is any real constant. Now we’re told that we want to find this antiderivative by considering the product rule. We recall that the product rule for two differentiable functions 𝑒 and 𝑣 is given by 𝑒𝑣 prime is equal to 𝑒 prime 𝑣 plus 𝑒𝑣 prime. The derivative of the product 𝑒𝑣 is the derivative of 𝑒 multiplied by 𝑣 plus 𝑒 multiplied by the derivative of 𝑣.

Now, looking at the function 𝑓 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 𝑒 to the power of π‘₯ is itself. And so the general antiderivative of 𝑒 to the power of π‘₯ is 𝑒 to the power of π‘₯ plus 𝐢. Comparing the expression we’ve been given for 𝑓 prime of π‘₯ with the definition of the product rule, it appears as if one term in our product, either 𝑒 or 𝑣, is 𝑒 to the power of π‘₯ as it can take the role of an antiderivative in one term and its derivative in the other. So let’s assume for now that 𝑣 is equal to 𝑒 to the power of π‘₯.

If this is correct and we’ve assigned 𝑣 and 𝑣 prime the right way around, then this would mean that in the first term, 𝑒 prime is equal to one over root π‘₯. And in the second term, 𝑒 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‘Ž plus one over π‘Ž plus one is π‘₯ to the power of π‘Ž, where π‘Ž is a real constant not equal to negative one. And so it follows that the general antiderivative of π‘₯ to the power of π‘Ž, where π‘Ž is again not equal to negative one, is π‘₯ to the power of π‘Ž plus one over π‘Ž plus one plus the constant of antidifferentiation 𝐢.

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

So by comparing the expression for 𝑓 prime of π‘₯ with the product rule, we’ve found that this is satisfied if 𝑒 is equal to two root π‘₯. 𝑒 prime is then equal to one over root π‘₯, 𝑣 is equal to 𝑒 to the power of π‘₯, and 𝑣 prime is also equal to 𝑒 to the power of π‘₯. The function 𝑓 of π‘₯ is then the product of 𝑒 and 𝑣. That’s two root π‘₯ multiplied by 𝑒 to the power of π‘₯. And we must include at this stage a general constant of antidifferentiation 𝐢.

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

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