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Find $ y' $ and $ y". $ $ y = \frac {\ln x}{l …

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Problem 23 Medium Difficulty

Find $y^{\prime}$ and $y^{\prime \prime}$
$y=\sqrt{x} \ln x$

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October 27, 2020

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October 27, 2020

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Video Transcript

All right, our goal is to take the derivative of the function on the left and red y equals square root of x times. L n of x notice it's a product, so i can call this part f and this part g and just remind her on the far right, the product rule. Whenever i want to take the derivative of 2 functions that are multiplied in this case. F and g, then the answer is f: prime g plus g prime f. So let's go ahead and solve then for y prime y prime is the derivative of f. Let'S go ahead and rewrite root x as x to the 1 half so that we can do power rule okay. So our derisive then, is the derivative of f to 1 half x to the minus 1 half that is by power rule. Then we multiply by g, which is l n of x. Next, we add, take the derivative of g and the derivative of l. N x is just 1 over x and we still multiply by x to the 1 half, because that is our f okay. So we do need to take another derivative, maybe keeping this up a little bit. Let'S see if i rearrange this, i get l n of x over 2 square root of x, plus, let's see x, i 1 half over x. You can subtract exponents of that's like x to the 1 half minus 1 or to the minus a half or 1 over square root of x. So that's what we have so far. I suppose we could factor out a 1 over root x and just leave it a little bit cleaner. That gives us times l n of x over 2 plus 1 point. So i think that's as bad as pretty as we're going to get for y prime of x, okay, so our next goal is to find y double prime of x now notice that we have yet another product, so we're going to be producta again this time. This will be f and the parentheses part will be g and i'm going to go ahead and like before i'm going to go back to the form minus a half to make it easier. So this will be my f. So when we do power rule, we already know that it's a minus the half there as the exponent. So let's go ahead and do protectoral 1 more time. The derivative of f is minus 1 half x to the minus 3 halves because we subtract 1 from the exponent and then g comes next left. As is then, we add and we have the derivative of g. The derivative l n of x is 1 of her x and the 2 goes along for the ride and derivative 1 is 0 and then times x, the minus 1 half all right. All we have left to do is a little clean up and then we will have finished our problems. This is great. Okay, so let's see what we have here, let's go ahead and just to kind of see what we have. Let'S go ahead and distribute this left side over into both parts of the parentheses. So, let's see we will get minus l n of x over 4 x to the 3 has on the bottom. That'S cleaning up this first part this part here and then we'll do the second part, so that will be minus 1 half x to the minus 3 has well. I can put it on the bottom 1 if we do that, make it pretty. So this will be 1 over x to the 3 halves on the bottom. I get this term and that will be plus i do have a half. Then i have x to the minus a half minus 1- more that's minus 3 halves or cos of 3 halves on the bottom. So we actually get something really nice to cancel so notice that these 2 terms add to 0. So they are canceled. So we actually get a very nice compact, second derivative of minus l n of x over 4 x to the 2. Don'T want to stay there 3 halves. Let'S make sure the 2 is there come on to that see. I can get that 2 to right there. We go okay, so it looks like we have solved both the first and second derivative, so excellent. Hopefully that helped have an amazing d.

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James Stewart

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Michael Jacobsen

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Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals