Video Transcript
By considering the product rule, find a function 𝑓 so that 𝑓 prime of 𝑥 is equal to 𝑒 to the power of 𝑥 over root 𝑥 plus two 𝑒 to the 𝑥 multiplied by root 𝑥.
We’ve been asked to find an expression for a function 𝑓 given an expression for its derivative 𝑓 prime of 𝑥. 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 𝑥. This will be a function 𝑓 of 𝑥 plus 𝐶, where 𝐶 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 𝑢 and 𝑣 is given by 𝑢𝑣 prime is equal to 𝑢 prime 𝑣 plus 𝑢𝑣 prime. The derivative of the product 𝑢𝑣 is the derivative of 𝑢 multiplied by 𝑣 plus 𝑢 multiplied by the derivative of 𝑣.
Now, looking at the function 𝑓 prime of 𝑥, we see that it is the sum of two expressions, each of which involves some power of 𝑥 and an exponential term. We should recall that the derivative of 𝑒 to the power of 𝑥 is itself. And so the general antiderivative of 𝑒 to the power of 𝑥 is 𝑒 to the power of 𝑥 plus 𝐶. Comparing the expression we’ve been given for 𝑓 prime of 𝑥 with the definition of the product rule, it appears as if one term in our product, either 𝑢 or 𝑣, is 𝑒 to the power of 𝑥 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 𝑥.
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, 𝑢 prime is equal to one over root 𝑥. And in the second term, 𝑢 is equal to two root 𝑥. But does this work? Well, we need to check. Using laws of exponents, one over root 𝑥 can be written as 𝑥 to the power of negative one-half. And two root 𝑥 can be written as two 𝑥 to the power of one-half. Then recalling the power rule of differentiation, we know that the derivative of 𝑥 to the power of 𝑎 plus one over 𝑎 plus one is 𝑥 to the power of 𝑎, where 𝑎 is a real constant not equal to negative one. And so it follows that the general antiderivative of 𝑥 to the power of 𝑎, where 𝑎 is again not equal to negative one, is 𝑥 to the power of 𝑎 plus one over 𝑎 plus one plus the constant of antidifferentiation 𝐶.
Finding the general antiderivative of what we suspect 𝑢 prime to be, which is 𝑥 to the power of negative one-half, we have 𝑥 to the power of negative one-half plus one over negative one-half plus one plus 𝐶. Negative a half plus one is positive a half. And dividing by a half is equivalent to multiplying by two. So if 𝑢 prime is 𝑥 to the power of negative one-half, then its antiderivative 𝑢 is two 𝑥 to the power of one-half plus 𝐶. And this is indeed equal to what we suspect 𝑢 should be when 𝐶 is equal to zero.
So by comparing the expression for 𝑓 prime of 𝑥 with the product rule, we’ve found that this is satisfied if 𝑢 is equal to two root 𝑥. 𝑢 prime is then equal to one over root 𝑥, 𝑣 is equal to 𝑒 to the power of 𝑥, and 𝑣 prime is also equal to 𝑒 to the power of 𝑥. The function 𝑓 of 𝑥 is then the product of 𝑢 and 𝑣. That’s two root 𝑥 multiplied by 𝑒 to the power of 𝑥. And we must include at this stage a general constant of antidifferentiation 𝐶.
So by considering the product rule, we found the function 𝑓 of 𝑥 such that 𝑓 prime of 𝑥 is equal to 𝑒 to the power of 𝑥 over root 𝑥 plus two 𝑒 to the power of 𝑥 multiplied by root 𝑥 to be 𝑓 of 𝑥 equals two root 𝑥𝑒 to the power of 𝑥 plus 𝐶.
