# Deriatives problem

1. Sep 17, 2009

### livelife92

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

find y=d/dx(e^x.e^x^2.e^x^3.................e^x^n) at 0<=x<=1
at x=0.5
2. Relevant equations

3. The attempt at a solution

so y=e^(x+x2+x3....xn)
1-i tried taking natural log on both sides ,thn differentiate.
it does not help..neither does using the d/dx(uv) help....both gives
y= e(summation of i=1 to n of x^r).(1+2x+3x^2.....nx^(n-1) )
= e^(x(x^n-1)/x-1).summation of nx^(n-1).
i cant simplify thies any further.and it looks horrible when x=0.5

and does not give me the slightest feeling that its the correct solution.. the answer is (4e)..but i have been told to try till i reach the solution!!!!

2. Sep 17, 2009

### livelife92

3. Sep 17, 2009

### HallsofIvy

Staff Emeritus
I don't see why writing it as $y= e^{x+x^2+x^2+\cdot\cdot\cdot+x^n}$ wouldn't work. The derivative of that is, of course, $(1+ 2x+ \cdot\cdot\cdot+ nx^{n-1})e^{x+x^2+ \cdot\cdot\cdot+ x^n}$ and that is to be evaluated at x= 1/2.

$x+ x^2+ \cdot\cdot\cdot+ x^n$ is a geometric series except that it is missing the initial "1". Its sum is given by $(1- x^{n+1})/(1- x)$. And the derivative, term by term is just the derivative of that: $1+ 2x+ \cdot\cdot\cdot+ nx^{n-1}}= [(n+1)x^n(1- x)+(1- x^{n+1})]/(1- x)^2$. Both can be evaluated at x= 1/2.

The only difficulty appears to be that the result should be 4e for all n. It certainly is not. For example when n= 1, this is just $e^x$ which has derivative $e^x$- evaluated at x= 1/2 that is $e^{1/2}$. For n= 2, this is $e^{x+x^2}$ and its derivative is $(1+2x)e^{x+x^2}$ which is $2e^{3/4}$ when x= 1/2.

Are you sure you have stated the problem correctly? Is it not, "find the limit, as n goes to infinity, of the derivative of $e^{x+ x^2+ \cdot\cdot\cdot+ x^n}$"? That would be, not, 4e, but $4e^2$.