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) := $\displaystyle d^2y/dx^2$
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) := $\displaystyle d^2y/dx^2$
You'll need the Leibniz Rule, the product rule, and the chain rule to differentiate your y(x). What do you get?
Just in case a picture helps...
... where (key in spoiler) ...
Spoiler:
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!