Second Partials D817F4

1. **State the problem:** We need to find the four second partial derivatives of the function $$f(x,y) = \sqrt{6 + 3x^2 + 7y^2}$$ which are $$f_{xx}, f_{yy}, f_{xy},$$ and $$f_{yx}$$. 2. **Recall the function and rewrite it:** $$f(x,y) = (6 + 3x^2 + 7y^2)^{\frac{1}{2}}$$ 3. **Find the first partial derivatives:** - Partial derivative with respect to $$x$$: $$f_x = \frac{1}{2}(6 + 3x^2 + 7y^2)^{-\frac{1}{2}} \cdot 6x = \frac{6x}{2\sqrt{6 + 3x^2 + 7y^2}} = \frac{3x}{\sqrt{6 + 3x^2 + 7y^2}}$$ - Partial derivative with respect to $$y$$: $$f_y = \frac{1}{2}(6 + 3x^2 + 7y^2)^{-\frac{1}{2}} \cdot 14y = \frac{14y}{2\sqrt{6 + 3x^2 + 7y^2}} = \frac{7y}{\sqrt{6 + 3x^2 + 7y^2}}$$ 4. **Find the second partial derivatives:** - Second partial derivative with respect to $$x$$: $$f_{xx} = \frac{d}{dx} \left( \frac{3x}{\sqrt{6 + 3x^2 + 7y^2}} \right)$$ Use the quotient rule or product rule: $$f_{xx} = \frac{3 \sqrt{6 + 3x^2 + 7y^2} - 3x \cdot \frac{1}{2} (6 + 3x^2 + 7y^2)^{-\frac{1}{2}} \cdot 6x}{6 + 3x^2 + 7y^2}$$ Simplify numerator: $$= \frac{3 \sqrt{6 + 3x^2 + 7y^2} - 9x^2 / \sqrt{6 + 3x^2 + 7y^2}}{6 + 3x^2 + 7y^2}$$ Multiply numerator terms by $$\sqrt{6 + 3x^2 + 7y^2}$$ to get common denominator: $$= \frac{3(6 + 3x^2 + 7y^2) - 9x^2}{(6 + 3x^2 + 7y^2)^{3/2}} = \frac{18 + 9x^2 + 21y^2 - 9x^2}{(6 + 3x^2 + 7y^2)^{3/2}} = \frac{18 + 21y^2}{(6 + 3x^2 + 7y^2)^{3/2}}$$ - Second partial derivative with respect to $$y$$: $$f_{yy} = \frac{d}{dy} \left( \frac{7y}{\sqrt{6 + 3x^2 + 7y^2}} \right)$$ Similarly, $$f_{yy} = \frac{7 \sqrt{6 + 3x^2 + 7y^2} - 7y \cdot \frac{1}{2} (6 + 3x^2 + 7y^2)^{-\frac{1}{2}} \cdot 14y}{6 + 3x^2 + 7y^2}$$ Simplify numerator: $$= \frac{7 \sqrt{6 + 3x^2 + 7y^2} - 49y^2 / \sqrt{6 + 3x^2 + 7y^2}}{6 + 3x^2 + 7y^2}$$ Multiply numerator terms by $$\sqrt{6 + 3x^2 + 7y^2}$$: $$= \frac{7(6 + 3x^2 + 7y^2) - 49y^2}{(6 + 3x^2 + 7y^2)^{3/2}} = \frac{42 + 21x^2 + 49y^2 - 49y^2}{(6 + 3x^2 + 7y^2)^{3/2}} = \frac{42 + 21x^2}{(6 + 3x^2 + 7y^2)^{3/2}}$$ - Mixed partial derivatives $$f_{xy}$$ and $$f_{yx}$$: Calculate $$f_{xy} = \frac{\partial}{\partial y} f_x$$: $$f_x = \frac{3x}{\sqrt{6 + 3x^2 + 7y^2}} = 3x (6 + 3x^2 + 7y^2)^{-1/2}$$ Differentiating with respect to $$y$$: $$f_{xy} = 3x \cdot \left(-\frac{1}{2}\right) (6 + 3x^2 + 7y^2)^{-3/2} \cdot 14y = -\frac{21xy}{(6 + 3x^2 + 7y^2)^{3/2}}$$ By Clairaut's theorem, $$f_{yx} = f_{xy} = -\frac{21xy}{(6 + 3x^2 + 7y^2)^{3/2}}$$ 5. **Final answers:** $$f_{xx} = \frac{18 + 21y^2}{(6 + 3x^2 + 7y^2)^{3/2}}$$ $$f_{yy} = \frac{42 + 21x^2}{(6 + 3x^2 + 7y^2)^{3/2}}$$ $$f_{xy} = f_{yx} = -\frac{21xy}{(6 + 3x^2 + 7y^2)^{3/2}}$$