Find y'(x) and y''(x) when y is defined with an integral

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Homework Statement



For x > 0, define y(x) := x \int_0^{log x} \! \sqrt{1 + e^t} dt -(2/3)(1 + x)^{3/2}

Calculate y'(x) := dy/dx
and

y''(x) := d^2y/dx^2

Homework Equations





The Attempt at a Solution



would like to work through this with person/people please. Not sure were to start
 
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Firstly, the FTC and the product rule will help you with that x*I(x). The second part of the function is easy to differentiate.
 
Char. Limit said:
Firstly, the FTC and the product rule will help you with that x*I(x). The second part of the function is easy to differentiate.

what is the FTC? will he first need to solve the integral?
 
The FTC is the Fundamental Theorem of Calculus. You could solve the integral, I suppose, but it would involve two substitutions, the first being u=e^t.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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