Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**BarandaMan****Guest**

bobbym wrote:

Hi;

Can you differentiate this wrt Pi?

It is just a power rule.

So I think this would become nWYP^n-1Pi^-n+1?

Can we write that in a nicer way?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

Hi;

That is not correct but it is close. After we move all the constants to the left you are differentiating:

Please differentiate that right now.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

Offline

**BarandaMan****Guest**

bobbym wrote:

Hi;

That is not correct but it is close. After we move all the constants to the left you are differentiating:

Please differentiate that right now.

That is I think -nPi^-n+1?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

Please use brackets.

What is the power rule?

Differentiate that please.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

Offline

**BarandaMan****Guest**

bobbym wrote:

Please use brackets.

What is the power rule?

Differentiate that please.

Sorry: (-n)(Pi)^(-n+1)

Erm I do not know, I think that is just simply 2x once differentiated? Multiply across then subtract the one?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

(-n)(Pi)^(-n+1)

That is not correct. Just follow the formula.

Power rule:

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

Offline

**BarandaMan****Guest**

bobbym wrote:

(-n)(Pi)^(-n+1)

That is not correct. Just follow the formula.

Power rule:

Ok, thank you.

Is that not what I did with x^2? Differentiate it and you get 2x? because 2x^(2-1) = 2x^1 = 2x?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

Yes, you were right for x^2 but wrong for Pi^(-n) which means you are not using the power rule or you are using it incorrectly.

Now, please differentiate this using the power rule:

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

**BarandaMan****Guest**

bobbym wrote:

Yes, you were right for x^2 but wrong for Pi^(-n) which means you are not using the power rule or you are using it incorrectly.

Now, please differentiate this using the power rule:

AHHHH! I get this, I think! So it is -nPi^(-n-1) and before I was adding one!?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

That is correct! You have one more part to do.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

**BarandaMan****Guest**

bobbym wrote:

That is correct! You have one more part to do.

YAY! Thank you!

SO, in whole must become, nWYP^(n-1)Pi(-n-1)? (For the bit you asked about before? )

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

nWYP^(n-1)Pi(-n-1)?

You left out an exponentiation sign in there.

It is Pi^(-n-1).

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

**BarandaMan****Guest**

Ah yes thank you bobby, sleep eyes.

Now this term is complete? But when you posted a solution, you had (nWY((Pi/P)^(-n-1)))/P^2 ? How did you get from the differentiated thing to this for this specific term?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

There is one more term to be differentiated. When we simplify it all up we will see.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

**BarandaMan****Guest**

Ok great! Now the last term, want to differentiate (YPi((Pi/P))^(-n))/P. So for this I get (-nYPi(Pi/P)^(-n-1))/P, but I know this is wrong because there are two Pi's and I am only doing one of them, how does this work??

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

Hi;

We are not done yet, not by a long shot. Which term are we working on?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

**BarandaMan****Guest**

The last term!

I think when we differentiate it we get (-nYPi((Pi/P)^(-n-1)))/P?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

The third term has a P in the denominator. I left it out because we concentrated on the numerator. So we are not done with the third term.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

**BarandaMan****Guest**

bobbym wrote:

The third term has a P in the denominator. I left it out because we concentrated on the numerator. So we are not done with the third term.

I see, apologies for not making this clear: P and Pi are completely different things.... P could be Y or X for example! It just happens to represent something beginning with P! But Pi is what we want to differentiate with respect to!

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

We differentiated that when the real problem is this:

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

**BarandaMan****Guest**

OH, so hwo do we do it?

oh my god why is this so difficult for such a little step, can we please go through this.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

Since the P is a constant just take the answer you got and put it over P.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

**BarandaMan****Guest**

bobbym wrote:

Since the P is a constant just take the answer you got and put it over P.

Ok now I am just getting confused, so that answer I had was correct?

Ok let's just leave this I don't think I can solve this problem anymore. Thank you though.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,302

Hi;

Okay, but we are almost done with the whole problem and are done with the third term.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

Offline

**BarandaMan****Guest**

Is there anyone here who can actually help me by explaining this methodically and quickly?

I have exams approaching in January and tried to solve this problem weeks ago. I cannot do it. I do not have the time to spend another 2 weeks trying to be directed to a correct answer. I have memorised the solution so that will be fine (solution is posted), but I do not understand how to differentiate it. Please help me.