Partial Derivative of Integral

  • Thread starter Victor8108
  • Start date
  • #1

Homework Statement



Find df/dx, f(x,y)=integral of sqrt(1-t^3)dt from x^2 to x^3.

Since it is asking to find the derivative with respect to x,should I regard t as a constant?

Homework Equations





The Attempt at a Solution



I tried to find the antiderivative of the integral sqrt(1-t^3)dt and then differentiate with respect to x.

However, I'm stuck in finding the antiderivative of the function , I tried using substitution u=1-t^3, du=-3t^2 dt. But that doesnt seem to be going the right direction.
 

Answers and Replies

  • #2
1,015
70
The variable "t" is a dummy variable we are summing over: the value of the definite integral is not a function of t. To see this another way, apply the Fundamental Theorem of Calculus to write the integral as a difference of two functions of x. You do not actually need to find an expression for these two functions, as all you will need are their derivatives. For example, consider
[tex]f(x) = \int_0^{x^2} \sin(t^2) dt[/tex]
The Fundamental Theorem of Calculus tells us that we can write f(x) like this as well, for each value of x:
[tex]f(x) = F(x^2) - F(0)[/tex]
where F(t) is any function whose derivative is [itex]\sin(t^2)[/itex] at each point t in the closed interval between t = 0 and t = [itex]x^2[/itex]. If we try to take the derivative of f(x), we find the following when we apply the chain rule:
[tex]f'(x) = F'(x^2)(2x) - F'(0)(0)[/tex]
We don't need to know any particular version of F, we just need to know what its derivative at [itex]x^2[/itex] and 0 is. Since [itex]x^2[/itex] and 0 are in the closed interval, we know the value of the derivative of F at those points are [itex]\sin(x^4)[/itex] and sin(0), respectively. Thus we get
[tex]f'(x) = 2x\sin(x^4)[/tex]
once we simplify.
 
Last edited:
  • #3
For those wondering why the equation f(x,y) has no y, here is the complete function:

f(x,y)= (x^2)y+integral of sqrt(1-t^3)dt from x^2 to x^3.

I figured the first part(x^2)y so I didnt put it in the equation. sorry about the inconvenience.
 
  • #4
Ok, according to the fundamental theorem of calculus:

if g(x)= f(t)dt from a to x, then g'(x)=f(x)

applying this to my problem, then it becomes g'(x)= sqrt(1-x^9)-sqrt(1-x^6)
 
  • #5
6,054
391
The fundamental theorem states that if F(a, b) is the integral of f(x) from a to b, then, letting g(t) = F(a, t), g'(t) = f(t).
 
  • #6
1,015
70
Ok, according to the fundamental theorem of calculus:

if g(x)= f(t)dt from a to x, then g'(x)=f(x)

applying this to my problem, then it becomes g'(x)= sqrt(1-x^9)-sqrt(1-x^6)

That's absolutely right.
 

Related Threads on Partial Derivative of Integral

  • Last Post
Replies
6
Views
3K
Replies
2
Views
5K
  • Last Post
Replies
7
Views
2K
Replies
3
Views
6K
  • Last Post
Replies
0
Views
1K
  • Last Post
2
Replies
26
Views
3K
  • Last Post
Replies
5
Views
17K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
989
  • Last Post
Replies
10
Views
797
Top