Question

Hi, can someone please help me obtain the second derivative of: y=x^2 + x^(1/2)? I get the first as 2X + (1/2)X^(-1/2), but I am stumped on the 2nd. Thanks.

Answer 1

The first is indeed right, the second is this: 2+(1/2)*(-1/2)x^(-3/2)=2-(1/4)x^(-3/2)

Answer 2

\[f(x)=x^2+\sqrt{x}\]
\[f'(x)=2x+\frac{1}{2\sqrt{x}}\]
\[f''(x)=2 -\frac{1}{4\sqrt{x^3}}\]

Answer 3

what thomas said. keep using power rule, in particular for
\[\frac{1}{2\sqrt{x}}=\frac{1}{2}x^{-\frac{1}{2}}\]

Answer 4

so
\[\frac{d}{dx}\frac{1}{2}x^{-\frac{1}{2}}=-\frac{1}{2}\times \frac{1}{2}x^{-\frac{1}{2}-1}\]

Answer 5

\[=-\frac{1}{4}x^{-\frac{3}{2}}=-\frac{1}{4\sqrt{x^3}}\]\]

Answer 6

thanks for the help. this was a multiple choice question given by my Prof. for extra credit and the choices are:
a)[(2X ^{3/2}-1)/ x^{3/2}]
b)[(8x^{3/2}+1)/4x^{3/2}]
c)[(8x ^{3/2}-1)/4x ^{3/2}]
d)[(2x ^{3/2}+1)/x ^{3/2}]
Thanks again.