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Suppose that k is a positive integer constant. Let f(x) = 1/(x^k-1). The nth derivative of this function is of the form f^(n) (x) =Pn(x)/(x^k -1)n+1 where Pn(x)is s...

Question

Suppose that k is a positive integer constant. Let f(x) = 1/(x^k-1). The nth derivative of this function is of the form f^(n) (x) =Pn(x)/(x^k -1)n+1 where Pn(x)is some polynomial function. it is easy to see thatP0(x) = 1 since we have differentiated zerotimes in thiscase. Also P1(x) = -kxk-1 a) suppose n>= 1. Find a formula for Pn(x)which depends only on x, n, k, Pn-1(x), andPn-1'(x).b) Find the value of P₁₀₀₁(1). Explain how youarrived at your answer.

Suppose that k is a positive integer constant. Let f(x) = 1/(x^k -1). The nth derivative of this function is of the form f^(n) (x) = Pn(x)/(x^k -1)n+1 where Pn(x) is some polynomial function. it is easy to see that P0(x) = 1 since we have differentiated zerotimes in this case. Also P1(x) = -kxk-1 a) suppose n>= 1. Find a formula for Pn(x) which depends only on x, n, k, Pn-1(x), and Pn-1'(x). b) Find the value of P₁₀₀₁(1). Explain how you arrived at your answer.



Answers

Let $k$ be a fixed positive integer. The nith derivative of $\frac{1}{x^{k}-1}$ has the form $\frac{P_{n}(x)}{\left(x^{k}-1\right)^{n+1}}$ where $P_{n}(x)$ is a polynomial. Find $P_{n}(1)$

So when this problem, we're asked to determine the derivative of this function Y equals Ellen of K X, where Kay is a constant and were asked to prove that why Prime is equal to one over X who are asked to prove that you were asked to prove that in two different ways. One using the chain rule and one using algebra, properties of the log, a rhythm and just differentiation. So first, let's use the chain rule. So why prime? We know that the derivative of Ellen of X there's one over X So the chain will tells us that the derivative of this function, Ellen F. K X, will be won over the argument of the law algorithm. But this argument is itself a function of X already. So we need to multiply by the derivative of this guy, which is just K because it's linear, right? It's some constant Times X well won over K X. Times K is just one over X, so that's proof Number one. Using the chain will prove Number two is to use products to use the product rule and differentiation. So it's rewrite why, and there's a rule in algebra in your algebra class. That tells us that when we have the log a rhythm of a product, we can rewrite that as the sun of the log, a rhythm of each of the factors. So I can write this as ln of K plus Ellen of X. So the law algorithm of K Times X is the son of the log of K and the log of X. So now when I take the derivative, I'm really just taking the derivative of each of these summons. And we know that the derivative of a sum is just the sum of the derivatives. In other words, I can take this guy's derivative and added to this guy's driven it Well, the derivative Ellen of K is just zero because Ellen of Kay is a constant right, Kay was assumed to be a constant. So the log of case constant and the Ellen of X is just one over X using this well known rule over here. So at the end of the day, we see that wide prime equals one over X, and that's the second proof

Yeah, Problem # 28. We're dealing with rational expressions and being able to do derivatives with the quotient rule. So applying the quotient rule here, F prime of X is going to be the derivative of the numerator, which is a times the denominator, C x plus k minus the numerator. A X plus B times the derivative of the denominator, which is C. All of this over the denominator. Yeah, squared. And now let's just see if we can do it without making mistakes and the simplification. So this is a C X plus A Okay minus a. See x minus B. See all of this over C X plus K quantity squared. What we see happening here is act X minus x. And so what this leaves me with, final answer is a k minus Bc Over c X plus k. Yes quantity squared.

So you have this one. I was gonna use the same trick that we did previously, right? This one is extra power 1/2. Just watch the tutorial just before this one did the same similar thing, right? This one is exit Apollo 1/2. This one is exited part of one. So if you multiply and you could just add the exponents exploiting here is one have explaining here is one. So one plus one have is 3/2 and then plus Exoo 1/2 because this is one right, So excellent. 1/2 times one is still excellent won't have. And from there, the derivative is pretty simple. This one through over two years in a drop down here says in every 32 x rate, and then we're gonna take away one me. So 3/2 minus warm. That is one of two. Great. So this one or two bliss this one have is gonna drop down here, so this is gonna be 1/2 exit. Take away one 1/2 minus one is negative. 1/2 here. We have a negative power, so we're gonna let it drop down. Right? So this is gonna be excellent. 1/2 then plus one over to right. This native 1/2 is gonna be to the power positive 1/2 which is the same as squarely of x ray. So, uh, actually, this power 1/2 a swell is not be screwed of X. A silly me. Right? That one. A swell. Yep. So you have those ass year answer, right?


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