Reciprocal Rule f(x) = 1/((x+1/x)^2) Welcome back. We’re going to do another reciprocal rule problem. This one is going to be a little bit more complicated and we’re usually going to use chain rule at the same time. How exciting. I’m going to go ahead and write our reciprocal rule just so that we have it in front of us. It is 1/f(x) = -f1 (x)/ [f (x)2]. All this is saying is that if you have a function with the one on the top and a function on the bottom that the derivative of it is the following—the derivative on top, the function squared on the bottom and then your sign now in front. So the problem we’re going to do f(x) = 1/(x + 1/x)2 so we know we can apply reciprocal rule to this problem because we have one on the top and we’ve got some function on the bottom. So we’ll do this one step at a time. We will go ahead and say, f1(x) the derivative equals—the first thing we need to do, we’re going to say negative because we know that we need the negative and then the top and bottom here I’m going to go ahead and write out the bottom because we already know what it is, it’s the bottom of this function squared so we’ll say x + 1/x2 is the bottom but then we need the square up here. So we add an extra set of brackets and square it again. So we basically leave it as it is and square it for the bottom. So really, all we have to do is deal with the top and that’s taking the derivative of the bottom right here. So to take the derivative of this, this is where chain rule is going to come in and what we’re going to do is apply chain rule so we’ll say, 2 × x + 1/x, I took the exponent on the 2 and I brought it out in front right here to this position because we’re treating everything inside the brackets here as one term bringing the two out in front multiplying exponent by the coefficient which is that implied 1 right there so we got the 2 out in front and then subtract 1 from the exponent 2 – 1 is 1 so that’s a 1 right there but of course it’s implied so we don’t need to write it. So we use chain rule to do that but then on top of it we have to multiply this whole function by the derivative of the inside, of this part right here. So we need to take the derivative of the inside as a second step. So the derivative of the inside we’ll do term by term, the derivative of x is 1 so let me go ahead and say—we’ll multiply here. The derivative of x is 1 and then plus the derivative of 1/x is actually –x/2 in a way that we do that I actually do it in my head I bring the x, I take this from right here and I convert this from 1/x to x-1 because when you have something in the denominator you can move it to the numerator and just flip the sign on the exponent from a positive to a negative so I take—I make this x-1 instead and then I bring that -1 out to the front so I say, -x and then to -2. This is –x-2 so that ends up converting to -1/x2 because just as I moved this from the denominator to the numerator and flip the exponent I can move this x-2 from the numerator to the denominator and flip the exponent from a negative to a positive. So I brought it from the bottom to the top and I’m bringing it from the top to the bottom and I have to flip the sign in the exponent both times. So this ends up being -1/x2 so this will be -1/x2 and since we have plus and then a negative we can just go ahead and make this minus. So now that we have multiplied—we multiply this by the derivative of the inside, the derivative of this section we’ve taken the full derivative of this whole function. So we can—you could simplify this but there’s nothing glaringly unacceptable on this so we can go ahead and put it up here and say, 2 × x + 1/x and then 1 – 1/x2 and that is the top. So this here is the derivative of the bottom and this is the bottom squared and then of course we have our negative sign at in front here you can’t forget that that’s the third components of the reciprocal rule. So there you have it, see you.