# Homework Help: Evaluating derivative of an equation

1. Nov 6, 2012

### Exizzle

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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.

2. Nov 6, 2012

### SammyS

Staff Emeritus
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))]

What is y(1) ?

3. Nov 6, 2012

### Exizzle

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?

4. Nov 6, 2012

### SammyS

Staff Emeritus
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.

5. Nov 7, 2012

### Exizzle

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!