Understanding Differentiation with x: ln and Exponents

  • Thread starter Thread starter M.Qayyum
  • Start date Start date
  • Tags Tags
    Differentiate
M.Qayyum
Messages
13
Reaction score
0

Homework Statement



1) ln { (x^2+x+1) / ( x^2-x+1) } here ln = natural log
2) x^2+x^(2y)=c here c=constant


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Show us your attempt at a solution.
 
Mark44 said:
Show us your attempt at a solution.

My attempt at 2)

x^2+x^(2y)=c

x^2+2ylnx=c

diff wrt x

2+(2lnx+1/x.2y.y')=0

1/x.2y.y'=-2-2lnx

y'= -2(lnx+1)x / 2y

is this correct and i can't understand the question no. one and how to start it's differentiation.
 
M.Qayyum said:
My attempt at 2)

x^2+x^(2y)=c

x^2+2ylnx=c
x2y \neq 2y ln(x).

You can write x2y as (eln(x))2y = e2y ln(x), and then differentiate.

For #1, ln(a/b) = ln(a) - ln(b). Use that property of logs before trying to differentiate.
M.Qayyum said:
diff wrt x

2+(2lnx+1/x.2y.y')=0

1/x.2y.y'=-2-2lnx

y'= -2(lnx+1)x / 2y

is this correct and i can't understand the question no. one and how to start it's differentiation.
 
Thank u very much...it really helps
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top