Differentiation problem

**oh507** · November 1st 2012, 02:50 AM

ax₁^a-1x₂^1-a/(1-a)x₁^ax₂^-a= p₁/p₂Hey guys. Stuck with this problem as I have forgotten my differentiation rules. I want to be able to show how to cancel down the equation so that it reads

a/(1-a) . x₂/x₁ = p₁/p₂

It would be great if someone could give me a step by step breakdown and remind me which rules I need to use here. Thank you

(sorry I didnt know how to use the mathematical script)

**TheEmptySet** · November 1st 2012, 04:36 AM

Originally Posted by oh507

ax₁^a-1x₂^1-a/(1-a)x₁^ax₂^-a= p₁/p₂Hey guys. Stuck with this problem as I have forgotten my differentiation rules. I want to be able to show how to cancel down the equation so that it reads

a/(1-a) . x₂/x₁ = p₁/p₂

It would be great if someone could give me a step by step breakdown and remind me which rules I need to use here. Thank you

(sorry I didnt know how to use the mathematical script)

It is not a derivative rule you need, but the law of exponents.

$\frac{a^{m}}{a^n}=a^{m-n}$ and $a^{-m}-\frac{1}{a^m}$

So you have

$\frac{ax_{1}^{a-1}x_{2}^{1-a}}{(1-a)x_1^{a}x_2^{-a}}=\frac{ax_{1}^{-1}x_2^{-1}}{(1-a)}=\frac{a}{(a-1)x_1x_2}$

**Assassin0071** · November 1st 2012, 04:37 AM

Hi oh507,

For this problem you don't need any differentiation rules. You need the rules about exponents to simplify the equation. Specifically you will need:

$x^{m} x^{n} = x^{m+n}$

and

$x^{-m} = \frac{1}{x^{m}}$

or

$\frac{1}{x^{-m}} = x^{m}$

to simplify your equation.

**oh507** · November 1st 2012, 04:51 AM

Ok, will relearn my rules and try figure out for myself. Thanks a lot!

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