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

Mar 8, 2016

$\frac{\text{d"}{"d} z}{\frac{{z}^{2} + 1}{\sqrt{z}}} = \frac{3 {z}^{2} - 1}{2 z \sqrt{z}}$

$\frac{\text{d"^2}{"d} {z}^{2}}{\frac{{z}^{2} + 1}{\sqrt{z}}} = \frac{3 \sqrt{z} \left({z}^{2} + 1\right)}{4 {z}^{3}}$

#### 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

• $u = {z}^{2} + 1$

$\frac{\text{d"u}{"d} z}{=} 2 z$

• $v = \sqrt{z}$

$\frac{\text{d"v}{"d} z}{=} \frac{1}{2 \sqrt{z}}$

So,

frac{"d"}{"d"z}((z^2 + 1)/sqrtz) = frac{vfrac{"d"u}{"d"z}-ufrac{"d"v}{"d"z}}{v^2}

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

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

Similarly, to find the second derivative, we have to change $u$ and $v$ to the numerator and denominator respectively of the first derivative.

• $u = 3 {z}^{2} - 1$

$\frac{\text{d"u}{"d} z}{=} 6 z$

• $v = 2 z \sqrt{z}$

$\frac{\text{d"v}{"d} z}{=} 3 \sqrt{z}$

So,

frac{"d"^2}{"d"z^2}((z^2 + 1)/sqrtz) = frac{"d"}{"d"z}(frac{"d"}{"d"z}((z^2 + 1)/sqrtz))

$= \frac{\text{d"}{"d} z}{\frac{3 {z}^{2} - 1}{2 z \sqrt{z}}}$

= frac{vfrac{"d"u}{"d"z}-ufrac{"d"v}{"d"z}}{v^2}

$= \frac{\left(2 z \sqrt{z}\right) \left(6 z\right) - \left(3 {z}^{2} - 1\right) \left(3 \sqrt{z}\right)}{{\left(2 z \sqrt{z}\right)}^{2}}$

$= \frac{12 {z}^{2} \sqrt{z} - \left(9 {z}^{2} \sqrt{z} - 3 \sqrt{z}\right)}{4 {z}^{3}}$

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