605: discussion15

Jie Zou

2021-12-10

ex 9-15 pg 711

find \(f_{x}, f_y, f_{xx}, f_{yy}, f_{xy}\) and \(f_{yx}\)

9

\[f(x,y) = x^2y+3x^2+4y-5\]


\[\begin{align*} f_x &= 2xy+6x \\ f_y &= x^2+4 \\ f_{xx} &= 2y+6 \\ f_{xy} &= 2x \\ f_{yx} &= 2x \\ f_{yy} &= 0 \end{align*}\]

10

\[f(x,y) = y^3 + 3xy^2 + 3x^2y + x^3\]


\[\begin{align*} f_x &= 3y^2+6xy+3x^2 \\ f_y &= 3y^2+6xy+3x^2 \\ f_{xx} &= 6y+6x \\ f_{xy} &= 6y+6x \\ f_{yx} &= 6y+6x\\ f_{yy} &= 6y+6x \end{align*}\]

11

\[f(x,y) = \frac{x}{y}\]


\[\begin{align*} f_x &= \frac{1}{y} \\ f_y &= -\frac{x}{y^2} \\ f_{xx} &= 0 \\ f_{xy} &= -\frac{1}{y^2} \\ f_{yx} &= -\frac{1}{y^2}\\ f_{yy} &= \frac{2x}{y^3} \end{align*}\]

12

\[f(x,y) = \frac{4}{xy}\]


\[\begin{align*} f_x &= -\frac{4}{x^2y} \\ f_y &= -\frac{4}{xy^2} \\ f_{xx} &= \frac{8}{x^3y} \\ f_{xy} &= \frac{4}{x^2y^2} \\ f_{yx} &= \frac{4}{x^2y^2}\\ f_{yy} &= \frac{8}{xy^3} \end{align*}\]

13

\[f(x,y) = e^{x^2+y^2}\]


let \(u = x^2+y^2\), \(\frac{du}{dx} = 2x, \frac{du}{dy} = 2y, e^u \frac{d}{du} = e^u\)

\[\begin{align*} f_x &= e^{x^2+y^2} \cdot 2x\\ f_y &= e^{x^2+y^2} \cdot 2y \\ f_{xx} &= (e^{x^2+y^2})'\cdot 2x + e^{x^2+y^2} \cdot(2x)' \\ &= e^{x^2+y^2} \cdot 4x^2 + e^{x^2+y^2} \cdot 2\\ f_{xy} &= e^{x^2+y^2} \cdot 2x \cdot 2y \\ f_{yx} &= (e^{x^2+y^2})'\cdot 2y + e^{x^2+y^2} \cdot(2y)'\\ &=e^{x^2+y^2} \cdot 2x \cdot 2y \\ f_{yy} &= (e^{x^2+y^2})'\cdot 2y + e^{x^2+y^2} \cdot(2y)' \\ &= e^{x^2+y^2} \cdot 4y^2 + e^{x^2+y^2} \cdot 2 \end{align*}\]

14

\[f(x,y) = e^{x+2y}\]


let \(u = x+2y\), then \(\frac{du}{dx}=1, \frac{du}{dy}=2, e^u\frac{d}{du} = e^u\)

\[\begin{align*} f_x &= e^{x+2y} \\ f_y &= e^{x+2y} \cdot2 \\ f_{xx} &= e^{x+2y} \\ f_{xy} &= e^{x+2y} \cdot2\\ f_{yx} &= e^{x+2y} \cdot2\\ f_{yy} &= e^{x+2y} \cdot 4 \end{align*}\]

15

\[f(x,y) = sin(x)cos(y)\]


\[\begin{align*} f_x &= sin'(x)cos(y) + sin(x)cos'(y) \\ &= cos(x)cos(y) \\ f_y &= -sin(x)sin(y) \\ f_{xx} &= cos'(x)cos(y) + cos(x)cos'(y) \\ &= -sin(x)cos(y) \\ f_{xy} &= -cos(x)sin(y) \\ f_{yx} &= -sin'(x)sin(y)+[-sin(x)]sin'(y) \\ &= -cos(x)sin(y)\\ f_{yy} &= -sin(x)cos(y) \end{align*}\]