How do you find the first and second derivatives of #f(z)= (z^2+1)/(sqrt (z))# using the quotient rule?

Question

Scarlett Allison · Accepted Answer

#frac{"d"}{"d"z}((z^2 + 1)/sqrtz) = frac{3z^2 - 1}{2zsqrtz}#

#frac{"d"^2}{"d"z^2}((z^2 + 1)/sqrtz) = frac{3sqrtz (z^2+1)}{4z^3}# The quotient rule: #frac{"d"}{"d"z}(u/v) = frac{vfrac{"d"u}{"d"z}-ufrac{"d"v}{"d"z}}{v^2}# In this question #frac{"d"u}{"d"z} = 2z# #frac{"d"v}{"d"z} = 1/{2sqrtz}# So, #frac{"d"}{"d"z}((z^2 + 1)/sqrtz) = frac{vfrac{"d"u}{"d"z}-ufrac{"d"v}{"d"z}}{v^2}# #= frac{sqrtz(2z)-(z^2 + 1)1/{2sqrtz}}{sqrtz^2}# #= frac{3z^2 - 1}{2zsqrtz}# Similarly, to find the second derivative, we have to change #u# and #v# to the numerator and denominator respectively of the first derivative. #frac{"d"u}{"d"z} = 6z# #frac{"d"v}{"d"z} = 3sqrtz# So, #frac{"d"^2}{"d"z^2}((z^2 + 1)/sqrtz) = frac{"d"}{"d"z}(frac{"d"}{"d"z}((z^2 + 1)/sqrtz))# #= frac{"d"}{"d"z}(frac{3z^2 - 1}{2zsqrtz})# #= frac{vfrac{"d"u}{"d"z}-ufrac{"d"v}{"d"z}}{v^2}# #= frac{(2zsqrtz)(6z)-(3z^2 - 1)(3sqrtz)}{(2zsqrtz)^2}# #= frac{12z^2sqrtz-(9z^2sqrtz - 3sqrtz)}{4z^3}# #= frac{3sqrtz (z^2+1)}{4z^3}#

Adam Adkins · Answer

undefined To find the first and second derivatives of $ f(z) = \frac{{z^2 + 1}}{{\sqrt{z}}} $ using the quotient rule:

1. First Derivative:
$$ f'(z) = \frac{{(z^2 + 1)' \cdot \sqrt{z} - (z^2 + 1) \cdot (\sqrt{z})'}}{{(\sqrt{z})^2}} $$

$$ = \frac{{(2z) \cdot \sqrt{z} - (z^2 + 1) \cdot \frac{1}{2}z^{-\frac{1}{2}}}}{{z}} $$

$$ = \frac{{2z\sqrt{z} - \frac{1}{2}(z^2 + 1)z^{-\frac{1}{2}}}}{{z}} $$

$$ = \frac{{2z\sqrt{z} - \frac{1}{2}z^{3/2} - \frac{1}{2}z^{-1/2}}}{{z}} $$

$$ = 2\sqrt{z} - \frac{1}{2}z^{1/2} - \frac{1}{2}z^{-1/2} $$

2. Second Derivative:
$$ f''(z) = \frac{{(2\sqrt{z})' \cdot z - 2\sqrt{z} \cdot 1 - \left(\frac{1}{2}z^{1/2} - \frac{1}{2}z^{-1/2}\right)' \cdot z - \frac{1}{2}z^{1/2} - \frac{1}{2}z^{-1/2} \cdot 1}}{{z^2}} $$

$$ = \frac{{\frac{1}{2}z^{-\frac{1}{2}} \cdot z - 2\sqrt{z} - \left(\frac{1}{4}z^{-\frac{3}{2}} + \frac{1}{4}z^{-\frac{3}{2}}\right) \cdot z - \frac{1}{2}z^{-\frac{1}{2}} - \frac{1}{2}z^{-\frac{1}{2}}}}{{z^2}} $$

$$ = \frac{{\frac{1}{2}z^{1/2} - 2\sqrt{z} - \frac{1}{2}z^{-1/2} - \frac{1}{2}z^{-1/2}}}{{z^2}} $$

$$ = \frac{{\frac{1}{2}z^{1/2} - 4\sqrt{z}}}{{z^2}} $$

$$ = \frac{1}{2}z^{-1/2} - \frac{4}{z^{3/2}} $$