\(\begin{array}{l}
\textbf{Bài 4:}\\
\quad f(x,y) = \ln(x^2 + 2y^2)\\
\text{Ta được:}\\
\bullet\quad \dfrac{\partial f}{\partial x}=\dfrac{2x}{x^2 + 2y^2}\Rightarrow \dfrac{\partial f}{\partial x}(1,2) = \dfrac29\\
\bullet\quad \dfrac{\partial f}{\partial y}=\dfrac{4y}{x^2 + 2y^2}\Rightarrow \dfrac{\partial f}{\partial y}(1,2)=\dfrac89\\
\textbf{Bài 5:}\\
\quad f(x,y) = y^2\ln4x\\
\text{Ta có:}\\
\bullet\quad \dfrac{\partial f}{\partial x} = \dfrac{y^2}{x}\\
\bullet\quad \dfrac{\partial f}{\partial y} = 2y\ln4x\\
\text{Khi đó:}\\
\quad df(1,1) = \dfrac{\partial f}{\partial x}(1,1)dx + \dfrac{\partial f}{\partial y}(1,1)dy\\
\Leftrightarrow df(1,1) = 1dx + 4\ln2dy\\
\textbf{Bài 6:}\\
\quad f(x,y) = 3x^2 - 2xy + 4y^2\\
\text{Ta có:}\\
\bullet\quad \dfrac{\partial f}{\partial x} = 6x- 2y\\
\bullet\quad \dfrac{\partial f}{\partial y} = -2x + 8y\\
\text{Khi đó:}\\
\quad \nabla f(1,1) =\left( \dfrac{\partial f}{\partial x}(1,1);\dfrac{\partial f}{\partial x}(1,1)\right)=(4;6)\\
\textbf{Bài 7:}\\
\quad f(x,y) = x^4y^3 - x^3y^4 + 5\\
\text{Ta có:}\\
\bullet\quad \dfrac{\partial f}{\partial x} = 4x^3y^3 - 3x^2y^4\\
\bullet\quad \dfrac{\partial f}{\partial y} = 3x^4y^2 - 4x^3y^3\\
\text{Khi đó:}\\
\bullet\quad \dfrac{\partial^2 f}{\partial x^2}= \dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial x}\right)=12x^2y^3 - 6xy^4\\
\bullet\quad \dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right)=12x^2y^2 - 12x^2y^3\\
\bullet\quad \dfrac{\partial^2 f}{\partial y^2}=\dfrac{\partial}{\partial y}\left( \dfrac{\partial f}{\partial y}\right)= 6x^4y - 12x^3y^2\\
\text{Ta được:}\\
\quad d^2f = \dfrac{\partial^2 f}{\partial x^2}dx^2 + 2\dfrac{\partial^2 f}{\partial x\partial y}dxdy + \dfrac{\partial^2 f}{\partial y^2}dy^2\\
\Leftrightarrow d^2f = (12x^2y^3 - 6xy^4)dx^2 + 2(12x^2y^2 - 12x^2y^3)dxdy + (6x^4y - 12x^3y^2)dy^2\\
\text{Bài 8:}\\
\quad f(x,y) = y\ln(xy)\\
\text{Ta có:}\\
\bullet\quad \dfrac{\partial f}{\partial x}=\dfrac yx\\
\bullet\quad \dfrac{\partial f}{\partial y}= \ln(xy) + 1\\
\text{Khi đó:}\\
\bullet\quad \dfrac{\partial^2f}{\partial x^2}= \dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial x}\right) = -\dfrac{y}{x^2}\\
\bullet\quad \dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right)= \dfrac1x\\
\bullet\quad \dfrac{\partial^2f}{\partial y^2}= \dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial y}\right) =\dfrac1y\\
\text{Ta được:}\\
\quad d^2f = \dfrac{\partial^2 f}{\partial x^2}dx^2 + 2\dfrac{\partial^2 f}{\partial x\partial y}dxdy + \dfrac{\partial^2 f}{\partial y^2}dy^2\\
\Leftrightarrow d^2f = -\dfrac{y}{x^2}dx^2 + \dfrac{2}{x}dxdy + \dfrac{1}{y}dy^2
\end{array}\)
