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

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Mar 8, 2016

#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}#

Explanation:

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

  • #u = z^2 + 1#

#frac{"d"u}{"d"z} = 2z#

  • #v = sqrtz#

#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.

  • #u = 3z^2 - 1#

#frac{"d"u}{"d"z} = 6z#

  • #v = 2zsqrtz#

#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}#

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