편도함수

(a) As $f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ is independent of $z$, we have $f_z(x,y,z) = 0$ and hence

$$f_{xyz}(x,y,z) =f_{zxy}(x,y,z) =0$$

(b) Write $u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$, $v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big)$ and $w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big)$. Then

• As $u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ is independent of $z$, we have $u_z(x,y,z) = 0$ and hence $u_{xyz}(x,y,z) =u_{zxy}(x,y,z) =0$
• As $v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big)$ is independent of $y$, we have $v_y(x,y,z) = 0$ and hence $v_{xyz}(x,y,z) =v_{yxz}(x,y,z) =0$
• As $w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big)$ is independent of $x$, we have $w_x(x,y,z) = 0$ and hence $w_{xyz}(x,y,z) =0$

As $f(x,y,z)=u(x,y,z)+v(x,y,z)+w(x,y,z)$, we have

$$f_{xyz}(x,y,z)=u_{xyz}(x,y,z)+v_{xyz}(x,y,z)+w_{xyz}(x,y,z)=0$$

(c) In the course of evaluating $f_{xx}(x,0,0)$, both $y$ and $z$ are held fixed at $0$. Thus, if we set $g(x) = f(x,0,0)$, then $f_{xx}(x,0,0)=g''(x)$. Now

$$g(x) = f(x,0,0) = \arctan\big(e^{\sqrt{xyz}}\big)\Big|_{y=z=0} =\arctan(1) =\frac{\pi}{4}$$

for all $x$. So $g'(x)=0$ and $g''(x)=0$ for all $x$. In particular,

$$f_{xx}(1,0,0) = g''(1) = 0$$