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If Vglx,y) = (2xy,xe+ 1+7) and g(0,1) = +6,then g(2,0) =here t0 search...

Question

If Vglx,y) = (2xy,xe+ 1+7) and g(0,1) = +6,then g(2,0) =here t0 search

If Vglx,y) = (2xy,xe+ 1+7) and g(0,1) = +6,then g(2,0) = here t0 search



Answers

If $ g(x) = x/e^x, $ find $ g^{(n)}(x). $

We have the function X divided by E to the X, and we want toe figure out what it's empty derivative is for any end. Well, the first thing I'm gonna do is I'm going to rewrite this as a product of two functions as opposed to a quotient of two functions. Because the product rule is a little bit simpler for derivatives and the quotient rule. And because I only have one term and the denominator, it's not too difficult to rearrange this quotient to be a product. Now, before we start taking derivatives, let's review the product rule. The product rule states that the derivative of F times G for two functions F N g is f prime G was f g prime. So if we want to figure out the end derivative, a good place to start is just by taking group. If we notice some sort of pattern, so buy the product rule. The derivative of X is one, so the first term is eat of the minus X, and the second term we get X times, the derivative of E to the negative X. Now that we're going to be using the chain rule because we have a composition of functions. We have Ito a function and then we have the inter function minus X and by the channel, the derivative of the outer function. E to the something is each of us something. And now I'm multiplied by the derivative of the inter function minus acts. The derivative of minus X is minus one. So I'm going to put a minus here instead of a plus. And so here's our first derivative. Well, one derivative probably won't show us an obvious pattern. So let's try the second derivative. We just took the derivative of each of the minus X. It was minus E to the minus X now minus. And now we want to take the derivative of X E to the minus X, which we just did. Why prime is that entire derivative? So I can just substitute that in here, and then I can distribute the negative to get minus E to the minus X minus e to the minus X, which is really minus to e to the minus X plus X e to the minus x Well, let's try the third derivative, the derivative of minus two e to the minus X. Well, we took that. We know that the derivative of E to the minus X is minus eat the minus X, and we have a constant of minus two being a factor here and the derivative of a constant times. The function is that constant times the derivative of the function. So we can keep the minus two right there multiplied by negative one, and we get to e to the minus X plus the derivative of X E to the minus X, which again is just why prime? So we get e to the minus x minus X e to the minus X, which simplifies to three e to the minus x minus X e to the minus x. So one thing we can notice is that why prime was e to the minus x minus X e to the minus X and why triple prime is three e to the minus X minus X e to the minus x so those do look a bit similar. If they're the same, except for the coefficient on the first term is equal to the degree of your derivative. But that seems to that seems like the so far the odd derivatives air following a pattern. Let's see if the even derivatives we're gonna follow a pattern. So if I take the derivative of this of white triple prime, I get minus three e to the minus x minus. Why prime which again is e to the minus X minus X e to the minus X, which is minus four e to the minus X plus X e to the minus x So now we see that why Double prime is minus two e to the minus X plus X E to the minus X and why quadruple prime is minus four, either the minus X plus X even minus sechs. So again it looks like the evens are also following a pattern. It's e to the minus. X plus XY. The minus minus X except the first term has the coefficient equal to the degree of your derivative. Now, is this just coincidence or it doesn't make sense? Is it actually true? Well, let's start with the odds. So we have y prime is e to the minus X minus X e to the minus X, and we also have why Triple Prime is three e to the minus X minus X eat of the minus sechs. Let's take a look at what happens when we take the derivative of why triple prime we do. We get minus three e to the minus X minus y prime, and that gets us our result. Now we see that that's the exact same thing as what we did for why prime, except for the fact that instead of a coefficient of three, we had a coefficient of one here. But if you look through the process as we take wide, double prime, that coefficient just sits there the whole time. And so the only difference in what happens when we took the derivative of y prime, as opposed to a why triple prime is what coefficient we subtracted. I'm sorry, what coefficient we add one to after negating it. Which means that for the odd power, the odd, uh, orders of derivatives this pattern does indeed hold that if it holds for some odd number, then we see by the the pattern of how we're taking this derivative that it will hold for the next number. And in the exact same way we can see that the even numbers are also going to follow this pattern that if you look at, um, why double prime minus two e to the minus sechs plus X eat of the minus X When we take that derivative, How is that? While actually to see how is that different from taking the derivative of y quadruple prime? Let's actually take the fifth, the fifth derivative. We get four e to the minus x plus why prime, which is e to the minus X minus X e to the minus X, which is five e to the minus X minus X e to the minus x. So again, you see that the procedure of taking the derivative of Y double prime is the same. Exact thing is taking the as taking the derivative of white quadruple crime except for the coefficient that sits there the whole time and doesn't influence anything except the final coefficient. It's what coefficient do you start with that you end up negating and adding one, too. And so that means that if this pattern holds for some, even number and we can start with 20 is a special case, um, then it will hold for the next one. So In fact, these patterns are, um, correct. Now we just have to figure out how to write them. So this is going to be a piece wise formula. So we can write that. Why the end derivative of why is so? Let's just start with f N a zero because that is a somewhat special case. Or actually, it still makes. We actually don't even need that. Let's just say we'll start with N is even and check if zero fits that. So our pattern was We get minus and excellence. I'm gonna give myself a little bit more space here if and is. Even then we get minus and e to the minus x plus x e to the minus X. If N is even so, is this still truth and zero because zero is an even number. Well, if n a zero, then we just get X e to the minus x so it does in fact hold. And for odd we get and e to the minus X minus X e to the minus x for and odd. And since odd and even numbers makeup well, non negative, odd and even numbers make up all non negative numbers numbers in this case meaning integers we're done

Yeah, For this problem, we want to first find the second derivative of s with respect to pee. So in this case, P is our variable t is treated as a constant. So to find the second derivative, we need to first find the first derivative. So as prime with respect to pee is going to equal to he times t because we are using the power rule for bringing this exponents down in front. And, um then we're just subtracting the exploded by one. So that's how we get to pee and then t is just are constant. And so the first derivative is two times p times t. But we want to find the second derivative with respect to pee. Hey, and that is just going to equal to t ain't because t is treated just like a constant. So this is the same thing as finding the derivative, for example, of five X, which would just be five. Okay, so the second derivative with respect to P is to t for part B. We want to find the second derivative with respect to t. So this time T is are variable and P is a constant, so the first derivative of s with respect to t it is just going to be p squared. Hey, because, um, tea is just being raised to the first power in our original equation. So if the first derivative with respect to t is just p squared, the second derivative with respect to T is zero because P is a constant. So peace, where would also be a constant and the derivative of a constant is zero? Yes. The second derivative with respect to T would be zero. Yeah.

Hello. So here we have our F. T. Is equal to E. To the negative tee times I had minus the natural log of T square times jihad. And we want to find here while the derivative of R F. T times the second derivative of R F. T. That's going to be equal to the R F T times the triple derivative here. So the third derivative of r f T plus the first derivative. Our prime ft times are double prime. So we're going to define our prime are double prime and our triple prime. So the first derivative here, our prime of teeth. Well that is going to be equal to we get a negative E. To the negative T. So a negative E. To the negative tee times I had. And then we get um the derivative of the next log of T squared is going to be again by the chain rule is going to be a minus to over tea time's jay hat. Okay, there's the first derivative. Then we find the second derivative. So it's our double prime. Um That is going to be equal to one of the derivative of the first derivative. Um is going to give us an E. To the negative tee times I had. And then um a minus a negative two over t squared times J hat. Okay. And then the third derivative um is going to give us well. Again here we get a negative E. To the negative tee times I had. And then we get the derivative of two over T squared. Again by the train rule, that's gonna be a minus four over t cubed time's jay hat. And then putting the values of R. T R prime are double prime are triple prime into our formula. Here we get well let's see. We get this is going to be equal, chew A um a negative E to the negative two T. Plus four over T cube times the natural log of T squared. And then we take this and then we're adding um a negative E. To the negative to T. And then minus four over T cubed. So um that is going to give us A negative two times. Eat to the negative to tea and then a -4 over tea cube plus four over T. Times The Natural Log of T squared.

What the derivative of G equals. When we said it equal to zero, we first need to find what g of X equals when we set it to zero. Let's start by setting this equation right here in for zero. If we plug in as you enter G, we don't know what it is because we don't know what it is. Plus muggins here for X. It's gonna be time. Zero sign of G 20 is equal to zero squared. Okay, now, let's all this out to find what G of zero equals. This is just gonna This is gonna cancel out because it's multiplied by zero and this is also just equal zero cost zero scored a zero. So this means g of zero is also equal to zero. Now, to find the derivative of G for zero, we can rewrite our equation in terms of Y and X g of G of X will equal why and the derivative of G of X will equal Why prime or a derivative of wife are we read it right here. Okay, Now we can solve our new equation. Using implicit differentiation, we can start by differentiating both sides of the equation in respect facts. Okay, I'll serve the left side of the dream of Why is it just gonna be one? But since your differential and respect to X, it's going to be why Prime, you know, we have a product rule will start with a group of X, which is this one. So it's gonna be just sign why? Plus our derivative of sign of why. Which is co sign of why. But since we're differentiating in respect to act, it's going to look like my wife prime again. This is all equal to two x are derivative of X squared. Okay. And now what we can do is isolate for our Why prime variable. So what I'm gonna do is just a subtract sign. Why from both sides and also factor out why prime from these two. Okay, that's factored out. And on the other side, we're gonna have to x minus sign of why. Okay. And I want to divide both sides by one plus x coast and why to solve for y prime. Okay. And now would that we've used implicit differentiation to solve for R Drood of of why or which is equal to our derivative of G. We can rewrite this all back in our original variables. Next is the same, but it's gonna be sign G of X all over one times x co sign G of X. And now we could do is plug in for zero because we have Oliver values as G of zero is also equal to zero and plug in zero for exits. Also zero. Was that in down here? Okay, Now, if we plug this end Ah, Sign of zero is equal to 21 zero. Sorry. It's able to zero. So on the top, we're just gonna have 220 which is zero minus zero. So it's all zero all over. One plus zero temps go, son of zero. So it's over one that is also with one. And this is your answer or the derivative of G when we plug it in for zero, that is also equal to zero. And that's your answer.


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