How can I differentiate this expression?

Click For Summary
SUMMARY

The discussion centers on the differentiation of an integral defined as H = ∫0f1(x) G(f2(ξ))G(f3(ξ))dξ. Participants clarify that since H is solely a function of x, the derivative dH/dy equals zero, provided that the limits of integration do not depend on y. If the limits were altered to f1(y), the differentiation would yield a non-zero result. The chain rule's application is debated, emphasizing that y's absence in the limits justifies the zero derivative.

PREREQUISITES
  • Understanding of univariate functions and their properties
  • Familiarity with integral calculus and differentiation techniques
  • Knowledge of the chain rule in calculus
  • Experience with function notation and limits of integration
NEXT STEPS
  • Study the application of the Fundamental Theorem of Calculus
  • Explore advanced differentiation techniques for integrals
  • Learn about Leibniz's rule for differentiating under the integral sign
  • Investigate the implications of variable limits of integration on differentiation
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on calculus, integral analysis, and differentiation techniques. This discussion is beneficial for anyone looking to deepen their understanding of the relationship between integration and differentiation.

cris(c)
Messages
21
Reaction score
0

Homework Statement


Consider three univariate distinct functions [itex]f_1(x),f_2(y),f_3(y)[/itex]. Let H be given by the following integral:

[itex]H=\int_{0}^{f_1(x)} G(f2(\xi))G(f3(\xi))d\xi[/itex]


The Attempt at a Solution


Then, computing dH/dy should give zero. However, I am not certain of this because the chain rule would give me:

[itex]\frac{dH}{dy}=\frac{\partial H}{\partial y} + \frac{\partial H}{\partial f_2}\frac{\partial f_2}{\partial y}+ \frac{\partial H}{\partial f_3}\frac{\partial f_3}{\partial y}[/itex]

and [itex]\frac{\partial H}{\partial f_j}\frac{\partial f_j}{\partial y}[/itex] are nonzero.
 
Physics news on Phys.org
cris(c) said:

Homework Statement


Consider three univariate distinct functions [itex]f_1(x),f_2(y),f_3(y)[/itex]. Let H be given by the following integral:

[itex]H=\int_{0}^{f_1(x)} G(f_2(\xi))G(f_3(\xi))d\xi[/itex]

The Attempt at a Solution


Then, computing dH/dy should give zero. However, I am not certain of this because the chain rule would give me:

[itex]\frac{dH}{dy}=\frac{\partial H}{\partial y} + \frac{\partial H}{\partial f_2}\frac{\partial f_2}{\partial y}+ \frac{\partial H}{\partial f_3}\frac{\partial f_3}{\partial y}[/itex]

and [itex]\frac{\partial H}{\partial f_j}\frac{\partial f_j}{\partial y}[/itex] are nonzero.
I had hoped one of our "experts" such as Dick, or micromass, or ... would have responded to this, but after looking at it a few times, I'll take a stab at it. (An informed stab at that.)

The integrand [itex]\displaystyle G(f_2(\xi))G(f_3(\xi))[/itex] is a function of ξ. You are integrating over ξ and the limits of integration are 0 and f1(x). Therefore, H is a function of x and only of x, so of course, [itex]\displaystyle \frac{dH}{dy}=0\,.[/itex]

If you intended the limits of integration to be 0 and f1(y), then finding [itex]\displaystyle \frac{dH}{dy}[/itex] makes sense.

If [itex]\displaystyle H=\int_{0}^{f_1(y)} G(f_2(\xi))G(f_3(\xi))d\xi\,,[/itex] the H(y) is the anti-derivative of [itex]\displaystyle G(f_2(\xi))G(f_3(\xi))[/itex] evaluated at ξ=f1(y), minus a constant.

The derivative of this result (w.r.t. y) should be pretty obvious, if this is indeed the question.
 
SammyS said:
I had hoped one of our "experts" such as Dick, or micromass, or ... would have responded to this, but after looking at it a few times, I'll take a stab at it. (An informed stab at that.)

The integrand [itex]\displaystyle G(f_2(\xi))G(f_3(\xi))[/itex] is a function of ξ. You are integrating over ξ and the limits of integration are 0 and f1(x). Therefore, H is a function of x and only of x, so of course, [itex]\displaystyle \frac{dH}{dy}=0\,.[/itex]

If you intended the limits of integration to be 0 and f1(y), then finding [itex]\displaystyle \frac{dH}{dy}[/itex] makes sense.

If [itex]\displaystyle H=\int_{0}^{f_1(y)} G(f_2(\xi))G(f_3(\xi))d\xi\,,[/itex] the H(y) is the anti-derivative of [itex]\displaystyle G(f_2(\xi))G(f_3(\xi))[/itex] evaluated at ξ=f1(y), minus a constant.

The derivative of this result (w.r.t. y) should be pretty obvious, if this is indeed the question.


Thanks SammyS. I thought exactly the same as you. However, I still don't see why the chain rule would be invalidated in this case. I know that so long as y does not appear in the limits of integration the integral should not change with y, but why the chain rule doesn't appear to say the sameÉ
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
6
Views
1K
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
1K
Replies
21
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K