Using logarithmic differentiation

rteng
Messages
26
Reaction score
0
MATH.jpg


using logarithmic differentiation I can get it to a point where I have...

ln(y)=20[ln(x+sqrt(x))-ln(x^2-2^x)]

I do not know what to do at that point
 
Physics news on Phys.org
just take the derivative as you normally would

also you will have \frac{y'}{y}= ...

bring Y to the other side and plug in what Y is. the rest is straight forward differentiation.
 
yeah but then you have a term that is:

[1/(x^2-2^x)]*(2x-d/dy2^x)

thats where I get stuck
 
i'm not really following

you mean on the x^{2}-2^{x} part? how would you take it's derivative?
 
y=((x+sqrt(x))/x^{2}-2^{x})^{20}

lny=20(ln(x+sqrt(x))-ln(x^{2}-2^{x}))

y'/y=(20/(x+sqrt(x))*(1+1/2(x)^{-1/2}))-(20/x^{2}-2^{x})*(2x-derivative of 2^x)
 
yes that part sorry
 
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...

Similar threads

Back
Top