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

Calculate $\displaystyle y'(x) := dy/dx$

and

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

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- Feb 21st 2011, 05:11 AMmaximus101find y'(x) and y''(x) when y is defined with an integral
For x > 0, define $\displaystyle y(x) := x$ $\displaystyle \int_0^{log x} \! \sqrt{1 + e^t} dt -(2/3)(1 + x)^{3/2} $

Calculate $\displaystyle y'(x) := dy/dx$

and

y''(x) := $\displaystyle d^2y/dx^2$ - Feb 21st 2011, 05:13 AMAckbeet
You'll need the Leibniz Rule, the product rule, and the chain rule to differentiate your y(x). What do you get?

- Feb 21st 2011, 06:38 AMHallsofIvy
Since there is no x inside the integral, strictly speaking you don't need the full "Leibniz rule", just the "Fundamental theorem of Calculus" and the chain rule.

- Feb 21st 2011, 08:01 AMtom@ballooncalculus
Just in case a picture helps...

http://www.ballooncalculus.org/draw/ftc/two.png

... where (key in spoiler) ...

__Spoiler__:

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Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

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