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# Find the derivative of the function.$y = (x + \frac {1}{x})^5$

## $\frac{5\left(x^{2}+1\right)^{4}\left(x^{2}-1\right)}{x^{6}}$

Derivatives

Differentiation

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##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

here we have a composite function, one function inside another. So we're going to use the chain rule. And before we start, I always like to write one over X as X to the negative first power. It's easier to differentiate that way. So here's our function. Now let's find the derivative. So we start with the derivative of the outside function. So we bring down the five and we raised the inside to the fourth. Now we need to find the derivative of the inside function, so we multiply by it. So the derivative of X is one, and then the derivative of X to the negative. First would be negative one times X to the negative second. Okay, now we just need to simplify this. So what I'm going to do is I'm going to rewrite my exes that have negative exponents as, ah, fractions. So we have five times x plus one over X to the fourth power times one minus one of Rex. What squared? Now that answer looks great to me, but what if you're told that you're not allowed to leave the fractions in there? So what can you do to alleviate that situation? I remember that simplifying is a bit of a judgment call, and some instructors will expect more simplifying than others. So what we could do is we could multiply this whole quantity by X to the sixth over X to the sixth, and I'll tell you why. So, for the first quantity that has fractions in it, we're going to need to multiply it by X to the fourth power. And for the second quantity that has fractions in it, we're going to need to multiply it by X to the second power. And if we do that, then we have to balance this out by dividing everything by X to the fourth annex to the second. So that would be X to the six. Okay, the reason we want to multiply this one by the X to the fourth power is so that we can combine them into five times X to the fourth. Okay, I'm gonna back up a second and go over a property with you before I do this. So remember this property. If you have a to a power, times be to a power, it's equivalent to a times B to the power. So if you had a to the fourth times be to the fourth. It would be equivalent to a times B to the fourth. That's what we're talking about here. So I'm gonna write this as X Times X plus one over X to the fourth. All right. So both of them are inside the fourth power now, and I'm going to do the same thing with the other one. I'm going to write it as well. We really don't have to. With the other one, we can leave it as it waas just X squared times one minus one over X squared. Okay, so now the next thing we're going to do is multiply the X by the X and the X by the one rex were distributing the X and then we're going to multiply. The X squared by the one and the X squared by the one over export were distributing the X squared. So we have five times a quantity X squared plus one to the fourth power times a quantity X squared minus one all over X to the sixth. So if your goal was to not only differentiate but simplify so that you no longer have those fractions in there, then that would be your answer. If you didn't need to do that, then you probably would have stopped at a previous step.

Oregon State University

Derivatives

Differentiation

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