# Calculus 2 second derivative of integral

1. Jan 25, 2014

### californicate

1. The problem statement, all variables and given/known data
Find d2/dx2 0x (1sint√(1+u^4)du)dt

2. Relevant equations

3. 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: Jan 25, 2014
2. Jan 25, 2014

### pasmith

Your solution is correct. If it helps, define
$$f(t) = \int_1^{\sin t} \sqrt{1 + u^4}\,du.$$
Then
$$\frac{d}{dx} \int_0^x \int_1^{\sin t} \sqrt{1 + u^4}\,du\,dt = \frac{d}{dx} \int_0^x f(t) \,dt = f(x)$$
and
$$\frac{df}{dx} = \frac{d}{dx} \int_1^{\sin x} \sqrt{1 + u^4}\,du = \cos x \sqrt{1 + \sin^4 x}$$