Problem 46 Easy Difficulty

Use logarithmic differentiation to find the derivative of the function.

$ y = (\sqrt x)^x $

Answer

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00:50

Frank L.

01:06

Doruk I.

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Video Transcript

in this problem. We are practicing, taking the derivative of a function using log arrhythmic differentiation, and that's going to make finding the derivative of the function were given much easier than a different way that we might use, like the limit definition or chain rule. So we're given the function. Why equals the square root of X rays to the X. Now we can simplify this by saying that this is the same thing is saying X rays to X over to. So the first thing that we're going to do is we'll take the natural log of each side and this is going to help us because we can move the exponents very easily using log properties. So we'll take the natural log of each side. We'll get the natural log of y equals the natural log of X rays to X over to. So now we can bring out that X over to term to get the natural log of y equals X over two times the natural log of X. Now, this is really nice, because we can use product rule now so well, find the derivative on each side will have won over why times by prime equals one half times the natural log of X plus X over two times one over X And again we found this side by using product rule. Now we could simplify that to say that one over. Why times y prime equals one half times and natural log of X plus one half. And now we need why prime by itself. So we would multiply by Why? To get white prime equals y over two times one plus the natural log of X. Now the only thing we have left to do is plug in. Why we know what why is that is the function that we started with. So why prime would be equal to the square root of X raised the X over two times one plus the natural log of X and that is our derivative of the function. I hope that this problem helped you understand a little bit more about lag, arrhythmic differentiation and why we use it and how that you could go through the process of using log properties and differentiation rules. In this case, the product rule