Calculus 2 second derivative of integral

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
californicate
Messages
12
Reaction score
0

Homework Statement


Find d2/dx2 0x (1sint√(1+u^4)du)dt

Homework Equations


The Attempt at a Solution


Initially I treated this problem as the second derivative of a double integral and thus quickly found myself at the result cosx√1+sin4x, by the fundamental theorem of calculus. However I realized that the problem asks for the second derivative according to x, and now I think my solution is incorrect. Would the correct method be to solve the integral √(1+u^4)du, then do the derivatives?
 
Last edited:
Physics news on Phys.org
californicate said:

Homework Statement


Find d2/dx2 0x (1sint√(1+u^4)du)dt


Homework Equations





The Attempt at a Solution


Initially I treated this problem as the second derivative of a double integral and thus quickly found myself at the result cosx√1+sin4x, by the fundamental theorem of calculus. However I realized that the problem asks for the second derivative according to x, and now I think my solution is incorrect. Would the correct method be to solve the integral √(1+u^4)du, then do the derivatives?

Your solution is correct. If it helps, define
[tex] f(t) = \int_1^{\sin t} \sqrt{1 + u^4}\,du.[/tex]
Then
[tex] \frac{d}{dx} \int_0^x \int_1^{\sin t} \sqrt{1 + u^4}\,du\,dt<br /> = \frac{d}{dx} \int_0^x f(t) \,dt<br /> = f(x)[/tex]
and
[tex] \frac{df}{dx} = \frac{d}{dx} \int_1^{\sin x} \sqrt{1 + u^4}\,du<br /> = \cos x \sqrt{1 + \sin^4 x}[/tex]
 
  • Like
Likes   Reactions: 1 person