Câu 1:
$\quad z(x,y)= xy + ye^{\displaystyle{\dfrac{x+y}{x}}}$
Ta có:
$+)\quad z_x' =y - \dfrac{y^2}{x^2}e^{\displaystyle{\dfrac{x+y}{x}}}$
$+)\quad z_y' =x + \dfrac{x+y}{x}e^{\displaystyle{\dfrac{x+y}{x}}}$
Khi đó:
$\quad z_x' + \dfrac yxz_y'$
$=y - \dfrac{y^2}{x^2}e^{\displaystyle{\dfrac{x+y}{x}}} + \dfrac yx\cdot \left(x + \dfrac{x+y}{x}e^{\displaystyle{\dfrac{x+y}{x}}}\right)$
$= 2y + \dfrac{y(x+y)}{x^2}e^{\displaystyle{\dfrac{x+y}{x}}}$
$= 2y + \dfrac{y}{x}e^{\displaystyle{\dfrac{x+y}{x}}}$
$= \dfrac{2xy + ye^{\displaystyle{\dfrac{x+y}{x}}}}{x}$
$= \dfrac{xy + \left(xy + ye^{\displaystyle{\dfrac{x+y}{x}}}\right)}{x}$
$= \dfrac{xy + z}{x}$
Vậy đẳng thức đúng
Câu 2:
Sửa đề: $\left(\dfrac1x\right)'' = \dfrac{2}{x^3}$
Đặt $y = f(x)= \dfrac1x$
$TXD: D = \Bbb R \backslash\{0\}$
Ta có:
$+)\quad \Delta y = \dfrac{1}{x+\Delta x} - \dfrac1x = - \dfrac{\Delta x}{x(x + \Delta x)}$
Khi đó:
$y' = f'(x)= \lim\limits_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}= \lim\limits_{\Delta x \to 0}\left[- \dfrac{\Delta x}{x(x + \Delta x)\Delta x}\right]= - \dfrac{1}{x^2}$
$+)\quad \Delta y' = - \dfrac{1}{(x + \Delta x)^2} + \dfrac{1}{x^2}= \dfrac{\Delta x(2x + \Delta x)}{x^2(x + \Delta x)^2}$
Khi đó:
$y'' = f''(x)= \lim\limits_{\Delta x \to 0}\dfrac{\Delta y'}{\Delta x}= \lim\limits_{\Delta x \to 0}\dfrac{\Delta x(2x + \Delta x)}{x^2(x + \Delta x)^2\Delta x} = \dfrac{2}{x^3}$
Vậy $\left(\dfrac1x\right)'' = \dfrac{2}{x^3}$
