Evaluating derivative of an equation

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Homework Help Overview

The problem involves evaluating the derivative of a complex product of functions defined as \( y = (x^2+1)^{1/2} \cdot (x^3+1)^{1/3} \cdot (x^4+1)^{1/4} \cdots (x^{100}+1)^{1/100} \) at \( x = 1 \). Participants are exploring methods to differentiate this expression and evaluate it at the specified point.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Some participants attempt to use product and chain rules for differentiation but find the process complicated. Others explore using natural logarithms for implicit differentiation, noting that this approach seems more manageable but still leads to difficulties in completing the evaluation. There is discussion about the evaluation of \( y(1) \) and the complexity of summing the exponents when substituting \( x = 1 \).

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the best way to evaluate \( y(1) \). Some have noted that while they have made progress in differentiating, they are still uncertain about how to simplify the evaluation of \( y(1) \) and are seeking further clarification or alternative methods.

Contextual Notes

Participants mention constraints such as the requirement to show all work within a limited space, which adds pressure to find a concise solution. There is also a recognition of the potential complexity involved in summing the series of exponents when evaluating \( y(1) \).

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Homework Statement



y = (x^2+1)^1/2 * (x^3+1)^1/3 * (x^4+1)^1/4 * ... * (x^100+1)^1/100

y'(1) Evaluate exactly.

Homework Equations



The Attempt at a Solution



I'm not exactly sure what is needed to solve this, but I tried using product/chain rule but that doesn't end up nicely at all.

I also tried using natural logs/implicit differentiation which was better but still couldn't finish it:

y = 1/2ln(x^2+1) + 1/3ln(x^3+1)

dy/dx = y [1/2(2x/x^2+1) + 1/2(3x^2/x^3+1)...]

So basically dy/dx of all the terms on the right are 1/2 when x=1?
If so, then
dy/dx = y [99*1/2] = y[49.5]

But it says to evaluate exactly so i'd need to plug in y which makes things complicated again? Any help would be appreciated, sorry if I didnt' make this very clear.
 
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Exizzle said:

Homework Statement



y = (x^2+1)^1/2 * (x^3+1)^1/3 * (x^4+1)^1/4 * ... * (x^100+1)^1/100

y'(1) Evaluate exactly.

Homework Equations



The Attempt at a Solution



I'm not exactly sure what is needed to solve this, but I tried using product/chain rule but that doesn't end up nicely at all.

I also tried using natural logs/implicit differentiation which was better but still couldn't finish it:

y = 1/2ln(x^2+1) + 1/3ln(x^3+1)
That should be ln(y) = 1/2ln(x^2+1) + 1/3ln(x^3+1) + ... + 1/100ln(x^100+1)
dy/dx = y [1/2(2x/x^2+1) + 1/2(3x^2/x^3+1)...]
That should be dy/dx = y [1/2(2x/(x^2+1)) + 1/3(3x^2/(x^3+1)) + ... + 1/100(100x^99/(x^100+1))]
So basically dy/dx of all the terms on the right are 1/2 when x=1?
If so, then
dy/dx = y [99*1/2] = y[49.5]

But it says to evaluate exactly so i'd need to plug in y which makes things complicated again? Any help would be appreciated, sorry if I didnt' make this very clear.
But your result is correct.

What is y(1) ?
 
SammyS said:
That should be ln(y) = 1/2ln(x^2+1) + 1/3ln(x^3+1) + ... + 1/100ln(x^100+1)

That should be dy/dx = y [1/2(2x/(x^2+1)) + 1/3(3x^2/(x^3+1)) + ... + 1/100(100x^99/(x^100+1))]

But your result is correct.

What is y(1) ?

Yea I guess it was lazy on my part.

That's the problem, I can't figure out a way to find out y(1) easily..

I know that it plugging 1 in would result in 2^(1/2) * 2^(1/3) * 2^(1/4)... *2^(1/100) and I could just add the exponents but that would take too much effort unless I'm forgetting something that I probably learned way back?
 
Exizzle said:
Yea I guess it was lazy on my part.

That's the problem, I can't figure out a way to find out y(1) easily..

I know that it plugging 1 in would result in 2^(1/2) * 2^(1/3) * 2^(1/4)... *2^(1/100) and I could just add the exponents but that would take too much effort unless I'm forgetting something that I probably learned way back?
Isn't ak∙am∙an = ak+m+n ?

... so, what is 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/100 ?

I know. It's still pretty messy.
 
SammyS said:
Isn't ak∙am∙an = ak+m+n ?

... so, what is 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/100 ?

I know. It's still pretty messy.


Hmm that's what i meant but yea i forgot how to do that.. but ill try to figure it out. We get about a third of a page to show all our work so i was just thinking there was an easier way to solve this. thanks for your help!
 

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