Calculus 2 second derivative of integral

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SUMMARY

The discussion centers on finding the second derivative of the integral defined by the expression 0∫x (1∫sint√(1+u^4)du)dt. The correct approach involves defining the function f(t) = ∫1^{sin t} √(1 + u^4) du and applying the Fundamental Theorem of Calculus. The solution confirms that the first derivative is f(x) and the second derivative is calculated as cos(x)√(1 + sin^4(x)). This method clarifies the distinction between treating the integral as a double integral and correctly applying the derivatives with respect to x.

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californicate
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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?
 
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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
<br /> f(t) = \int_1^{\sin t} \sqrt{1 + u^4}\,du.<br />
Then
<br /> \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)<br />
and
<br /> \frac{df}{dx} = \frac{d}{dx} \int_1^{\sin x} \sqrt{1 + u^4}\,du<br /> = \cos x \sqrt{1 + \sin^4 x}<br />
 
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