Answered: H(p,r) = p Vp + 2r

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter5: Inner Product Spaces

Section5.CR: Review Exercises

Problem 14CR

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Question

Second partial derivatives Find the four second partial derivatives of the following functions.

Expert Solution

Step 1

Solution -

Given function -

$H (p, r) = p^{2} \sqrt{p + 2 r}$

1st second partial derivative -

$\frac{d^{2} H}{d r^{2}} = \frac{d}{d r} (\frac{d}{d r} p^{2} \sqrt{p + 2 r})$

$\frac{d^{2} H}{d r^{2}} = \frac{d}{d r} (p^{2} . \frac{1}{2 \sqrt{p + 2 r}} . 2)$

$\frac{d^{2} H}{d r^{2}} = p^{2} \frac{d}{d r} (\frac{1}{\sqrt{p + 2 r}})$

$\frac{d^{2} H}{d r^{2}} = p^{2} (\frac{(\sqrt{p + 2 r}) (0) - 1 . \frac{2}{2 \sqrt{p + 2 r}}}{{(\sqrt{p + 2 r})}^{2}})$

$\frac{d^{2} H}{d r^{2}} = (\frac{- p^{2}}{(p + 2 r) \sqrt{p + 2 r}}) = \frac{- p^{2}}{{(p + 2 r)}^{\frac{3}{2}}}$

Step 2

2nd second partial derivative -

$\frac{d^{2} H}{d p^{2}} = \frac{d}{d p} (\frac{d}{d p} p^{2} \sqrt{p + 2 r})$

$\frac{d^{2} H}{d p^{2}} = \frac{d}{d p} (p^{2} . \frac{1}{2 \sqrt{p + 2 r}} + \sqrt{p + 2 r} . 2 p)$

$\frac{d^{2} H}{d p^{2}} = \frac{d}{d p} (\frac{p^{2} + 4 p (p + 2 r)}{2 \sqrt{p + 2 r}})$

$\frac{d^{2} H}{d p^{2}} = \frac{d}{d p} (\frac{5 p^{2} + 8 p r}{2 \sqrt{p + 2 r}})$

$\frac{d^{2} H}{d p^{2}} = (\frac{2 \sqrt{p + 2 r} . (10 p + 8 r) - (5 p^{2} + 8 p r) . \frac{1}{\sqrt{p + 2 r}}}{{(2 \sqrt{p + 2 r})}^{2}})$

$\frac{d^{2} H}{d p^{2}} = (\frac{2 (p + 2 r) . (10 p + 8 r) - (5 p^{2} + 8 p r)}{4 (p + 2 r) \sqrt{p + 2 r}})$

$\frac{d^{2} H}{d p^{2}} = (\frac{20 p^{2} + 16 p r + 40 p r + 32 r^{2} - 5 p^{2} - 8 p r}{4 {(p + 2 r)}^{\frac{3}{2}}})$

$\frac{d^{2} H}{d p^{2}} = (\frac{15 p^{2} + 32 r^{2} + 48 p r}{4 {(p + 2 r)}^{\frac{3}{2}}})$

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