Answer :
To find the second partial derivatives of the function [tex]\( f(x, y) = \ln(9x^2 + y^2 + 9) \)[/tex], let's proceed step by step.
### First Partial Derivatives
1. First partial derivative with respect to [tex]\( x \)[/tex]:
[tex]\[ f_x = \frac{d}{dx} \left( \ln(9x^2 + y^2 + 9) \right) \][/tex]
Using the chain rule, we get:
[tex]\[ f_x = \frac{1}{9x^2 + y^2 + 9} \cdot \frac{d}{dx}(9x^2 + y^2 + 9) \][/tex]
[tex]\[ f_x = \frac{1}{9x^2 + y^2 + 9} \cdot 18x \][/tex]
[tex]\[ f_x = \frac{18x}{9x^2 + y^2 + 9} \][/tex]
2. First partial derivative with respect to [tex]\( y \)[/tex]:
[tex]\[ f_y = \frac{d}{dy} \left( \ln(9x^2 + y^2 + 9) \right) \][/tex]
Using the chain rule, we get:
[tex]\[ f_y = \frac{1}{9x^2 + y^2 + 9} \cdot \frac{d}{dy}(9x^2 + y^2 + 9) \][/tex]
[tex]\[ f_y = \frac{1}{9x^2 + y^2 + 9} \cdot 2y \][/tex]
[tex]\[ f_y = \frac{2y}{9x^2 + y^2 + 9} \][/tex]
### Second Partial Derivatives
1. Second partial derivative with respect to [tex]\( x \)[/tex]:
[tex]\[ f_{xx} = \frac{d}{dx} \left( \frac{18x}{9x^2 + y^2 + 9} \right) \][/tex]
We will use the quotient rule [tex]\(\left(\frac{a}{b}\right)' = \frac{a'b - ab'}{b^2}\)[/tex]:
[tex]\[ f_{xx} = \frac{(18)(9x^2 + y^2 + 9) - (18x)(18x)}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{xx} = \frac{18(9x^2 + y^2 + 9) - 324x^2}{(9x^2 + y^2 + 9)^2} \][/tex]
Simplifying the numerator:
[tex]\[ f_{xx} = \frac{162x^2 + 18y^2 + 162 - 324x^2}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{xx} = \frac{18y^2 - 162x^2 + 162}{(9x^2 + y^2 + 9)^2} \][/tex]
Factor out a common factor in the numerator:
[tex]\[ f_{xx} = \frac{18(-18x^2 + y^2 + 9)}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{xx} = \frac{18(-18x^2 + y^2 + 9)}{(9x^2 + y^2 + 9)^2} \][/tex]
2. Second partial derivative with respect to [tex]\( y \)[/tex]:
[tex]\[ f_{yy} = \frac{d}{dy} \left( \frac{2y}{9x^2 + y^2 + 9} \right) \][/tex]
Applying the quotient rule:
[tex]\[ f_{yy} = \frac{2(9x^2 + y^2 + 9) - (2y)(2y)}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{yy} = \frac{2(9x^2 + y^2 + 9) - 4y^2}{(9x^2 + y^2 + 9)^2} \][/tex]
Simplifying the numerator:
[tex]\[ f_{yy} = \frac{18x^2 + 2y^2 + 18 - 4y^2}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{yy} = \frac{18x^2 - 2y^2 + 18}{(9x^2 + y^2 + 9)^2} \][/tex]
Factor out a common factor in the numerator:
[tex]\[ f_{yy} = \frac{2(18x^2 - 2y^2 + 9)}{(9x^2 + y^2 + 9)^2} \][/tex]
3. Mixed second partial derivative [tex]\( f_{xy} \)[/tex]:
[tex]\[ f_{xy} = \frac{d}{dy} \left( \frac{18x}{9x^2 + y^2 + 9} \right) \][/tex]
Using the quotient rule:
[tex]\[ f_{xy} = \frac{18x \cdot \frac{d}{dy}(9x^2 + y^2 + 9) - \frac{d}{dy}(18x) \cdot (9x^2 + y^2 + 9)}{(9x^2 + y^2 + 9)^2} \][/tex]
Simplifying:
[tex]\[ f_{xy} = \frac{18x \cdot 2y - 18x \cdot (9x^2 + y^2 + 9)}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{xy} = \frac{36xy}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{xy} = \frac{-36xy}{(9x^2 + y^2 + 9)^2} \][/tex]
4. Mixed second partial derivative [tex]\( f_{yx} \)[/tex]:
By Clairaut's theorem (assuming the function is sufficiently smooth), the mixed partial derivatives are equal:
[tex]\[ f_{yx} = f_{xy} \][/tex]
Therefore:
[tex]\[ f_{yx} = \frac{-36xy}{(9x^2 + y^2 + 9)^2} \][/tex]
### Summary
The second partial derivatives are:
[tex]\[ f_{xx} = \frac{18(-18x^2 + y^2 + 9)}{(9x^2 + y^2 + 9)} \][/tex]
[tex]\[ f_{yy} = \frac{2(-2y^2 + 9x^2 + 9)}{(9x^2 + y^2 + 9)} \][/tex]
[tex]\[ f_{xy} = \frac{-36xy}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{yx} = \frac{-36xy}{(9x^2 + y^2 + 9)^2} \][/tex]
### First Partial Derivatives
1. First partial derivative with respect to [tex]\( x \)[/tex]:
[tex]\[ f_x = \frac{d}{dx} \left( \ln(9x^2 + y^2 + 9) \right) \][/tex]
Using the chain rule, we get:
[tex]\[ f_x = \frac{1}{9x^2 + y^2 + 9} \cdot \frac{d}{dx}(9x^2 + y^2 + 9) \][/tex]
[tex]\[ f_x = \frac{1}{9x^2 + y^2 + 9} \cdot 18x \][/tex]
[tex]\[ f_x = \frac{18x}{9x^2 + y^2 + 9} \][/tex]
2. First partial derivative with respect to [tex]\( y \)[/tex]:
[tex]\[ f_y = \frac{d}{dy} \left( \ln(9x^2 + y^2 + 9) \right) \][/tex]
Using the chain rule, we get:
[tex]\[ f_y = \frac{1}{9x^2 + y^2 + 9} \cdot \frac{d}{dy}(9x^2 + y^2 + 9) \][/tex]
[tex]\[ f_y = \frac{1}{9x^2 + y^2 + 9} \cdot 2y \][/tex]
[tex]\[ f_y = \frac{2y}{9x^2 + y^2 + 9} \][/tex]
### Second Partial Derivatives
1. Second partial derivative with respect to [tex]\( x \)[/tex]:
[tex]\[ f_{xx} = \frac{d}{dx} \left( \frac{18x}{9x^2 + y^2 + 9} \right) \][/tex]
We will use the quotient rule [tex]\(\left(\frac{a}{b}\right)' = \frac{a'b - ab'}{b^2}\)[/tex]:
[tex]\[ f_{xx} = \frac{(18)(9x^2 + y^2 + 9) - (18x)(18x)}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{xx} = \frac{18(9x^2 + y^2 + 9) - 324x^2}{(9x^2 + y^2 + 9)^2} \][/tex]
Simplifying the numerator:
[tex]\[ f_{xx} = \frac{162x^2 + 18y^2 + 162 - 324x^2}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{xx} = \frac{18y^2 - 162x^2 + 162}{(9x^2 + y^2 + 9)^2} \][/tex]
Factor out a common factor in the numerator:
[tex]\[ f_{xx} = \frac{18(-18x^2 + y^2 + 9)}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{xx} = \frac{18(-18x^2 + y^2 + 9)}{(9x^2 + y^2 + 9)^2} \][/tex]
2. Second partial derivative with respect to [tex]\( y \)[/tex]:
[tex]\[ f_{yy} = \frac{d}{dy} \left( \frac{2y}{9x^2 + y^2 + 9} \right) \][/tex]
Applying the quotient rule:
[tex]\[ f_{yy} = \frac{2(9x^2 + y^2 + 9) - (2y)(2y)}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{yy} = \frac{2(9x^2 + y^2 + 9) - 4y^2}{(9x^2 + y^2 + 9)^2} \][/tex]
Simplifying the numerator:
[tex]\[ f_{yy} = \frac{18x^2 + 2y^2 + 18 - 4y^2}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{yy} = \frac{18x^2 - 2y^2 + 18}{(9x^2 + y^2 + 9)^2} \][/tex]
Factor out a common factor in the numerator:
[tex]\[ f_{yy} = \frac{2(18x^2 - 2y^2 + 9)}{(9x^2 + y^2 + 9)^2} \][/tex]
3. Mixed second partial derivative [tex]\( f_{xy} \)[/tex]:
[tex]\[ f_{xy} = \frac{d}{dy} \left( \frac{18x}{9x^2 + y^2 + 9} \right) \][/tex]
Using the quotient rule:
[tex]\[ f_{xy} = \frac{18x \cdot \frac{d}{dy}(9x^2 + y^2 + 9) - \frac{d}{dy}(18x) \cdot (9x^2 + y^2 + 9)}{(9x^2 + y^2 + 9)^2} \][/tex]
Simplifying:
[tex]\[ f_{xy} = \frac{18x \cdot 2y - 18x \cdot (9x^2 + y^2 + 9)}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{xy} = \frac{36xy}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{xy} = \frac{-36xy}{(9x^2 + y^2 + 9)^2} \][/tex]
4. Mixed second partial derivative [tex]\( f_{yx} \)[/tex]:
By Clairaut's theorem (assuming the function is sufficiently smooth), the mixed partial derivatives are equal:
[tex]\[ f_{yx} = f_{xy} \][/tex]
Therefore:
[tex]\[ f_{yx} = \frac{-36xy}{(9x^2 + y^2 + 9)^2} \][/tex]
### Summary
The second partial derivatives are:
[tex]\[ f_{xx} = \frac{18(-18x^2 + y^2 + 9)}{(9x^2 + y^2 + 9)} \][/tex]
[tex]\[ f_{yy} = \frac{2(-2y^2 + 9x^2 + 9)}{(9x^2 + y^2 + 9)} \][/tex]
[tex]\[ f_{xy} = \frac{-36xy}{(9x^2 + y^2 + 9)^2} \][/tex]
[tex]\[ f_{yx} = \frac{-36xy}{(9x^2 + y^2 + 9)^2} \][/tex]
