Question

How do you find the second derivative of y=x^2-(1/x)+3?

Answer 1

y' = 2x + x^-2
y'' = 2-2x^-3

Answer 2

\[y=x^2-x^{-1}+3\]
\[\delta y/\delta x=2x+x^{-2}\]
\[\delta ^2 y/\delta x^2=2-2x^{-3}\]

Answer 3

differentiate term by term
dy/dx = 2x - (-x^-2)
= 2x + x^-2
second derivative is found by a further differentiation:
d2y / dx^2 = 2 - 2x^-3
or 2 - 2/x^3

Answer 4

since the second derivative is just the derivative of the first derivative; id say you just take the derivative again ....