방향도함수와 기울기벡터

The partial derivatives, at a general point $(x,y,z)$ and also at the point of interest $(0,1,1)$, are

$$\begin{aligned} f_x(x,y,z)&=yz e^{xyz}\qquad & f_x(0,1,1)&= 1 \\ f_y(x,y,z)&=xz e^{xyz}\qquad & f_y(0,1,1)&= 0 \\ f_z(x,y,z)&=xy e^{xyz}\qquad & f_z(0,1,1)&= 0 \end{aligned}$$

So ${ \boldsymbol{\nabla} } f(0,1,1) =\left< 1,0,0\right>$ and the specified directional derivative is

$$\begin{aligned} D_{\frac{\left< 0,1,1\right>}{\sqrt{2}}}f(0,1,1) =\left< 1,0,0\right>\cdot\frac{\left< 0,1,1\right>}{\sqrt{2}} =0 \end{aligned}$$

In two dimensions, write $g(x,y) = y^2+\sin(xy)$. Then

$$\begin{aligned} { \boldsymbol{\nabla} } g =\left< g_x\,,\,g_y\right> =\left< y\cos(xy)\,,\, 2y + x\cos(xy) \right> \end{aligned}$$

In three dimensions, write $g(x,y,z) = y^2+\sin(xy)$. Then

$$\begin{aligned} { \boldsymbol{\nabla} } g =\left< g_x\,,\,g_y\,,\,g_z\right> =\left< y\cos(xy)\,,\, 2y + x\cos(xy)\,,\, 0 \right> \end{aligned}$$