Hello,
As somebody has distroyed my question, (may be he have not understood ?)
i will reply to y=(√x)+5
[tex]\displaystyle \boxed{\dfrac{d^n y}{dx^n} =(-1)^n*n!! *\dfrac{1}{2^n} *x^{-\frac{2n+1}{2} }\\}\\y'=\dfrac{1}{2} *x^{-\frac{1}{2}} \\\\y''=-\dfrac{1}{4} *x^{-\frac{3}{2}} \\...\\[/tex]
