Solving Derivative Problem for y=e^x.e^x^2.e^x^3....e^x^n at 0<=x<=1

  • Thread starter Thread starter livelife92
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on finding the derivative of the function y = e^(x + x^2 + x^3 + ... + x^n) at x = 0.5, where the expected answer is 4e. Participants explored various methods, including taking the natural logarithm and applying the product rule, but encountered difficulties in simplification. One contributor pointed out that the sum x + x^2 + ... + x^n resembles a geometric series, leading to a more straightforward derivative calculation. However, discrepancies arose regarding the expected result, suggesting a potential misstatement of the problem.

PREREQUISITES
  • Understanding of derivatives and differentiation rules
  • Familiarity with exponential functions and their properties
  • Knowledge of geometric series and their summation
  • Experience with evaluating limits and series as n approaches infinity
NEXT STEPS
  • Study the properties of geometric series and their derivatives
  • Learn about the application of the product rule in differentiation
  • Explore the concept of limits in calculus, particularly with infinite series
  • Investigate the behavior of exponential functions as parameters change
USEFUL FOR

Students in calculus, mathematicians exploring series and derivatives, and educators seeking to clarify concepts related to exponential functions and differentiation techniques.

livelife92
Messages
6
Reaction score
0

Homework Statement



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

Homework Equations





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 can't 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!
 
Physics news on Phys.org
and the answer also asks for a condtion.
 
livelife92 said:

Homework Statement



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

Homework Equations





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 can't 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!

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

[itex]x+ x^2+ \cdot\cdot\cdot+ x^n[/itex] is a geometric series except that it is missing the initial "1". Its sum is given by [itex](1- x^{n+1})/(1- x)[/itex]. And the derivative, term by term is just the derivative of that: [itex]1+ 2x+ \cdot\cdot\cdot+ nx^{n-1}}= [(n+1)x^n(1- x)+(1- x^{n+1})]/(1- x)^2[/itex]. 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 [itex]e^x[/itex] which has derivative [itex]e^x[/itex]- evaluated at x= 1/2 that is [itex]e^{1/2}[/itex]. For n= 2, this is [itex]e^{x+x^2}[/itex] and its derivative is [itex](1+2x)e^{x+x^2}[/itex] which is [itex]2e^{3/4}[/itex] 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 [itex]e^{x+ x^2+ \cdot\cdot\cdot+ x^n}[/itex]"? That would be, not, 4e, but [itex]4e^2[/itex].
 

Similar threads

Replies
7
Views
2K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
1
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K