题目
(5)z=lntan(x)/(y); (6)z=(1+xy)^y; (7)u=x^(y)/(z); (8)u=arctan(x-y)^z.
(5)$z=ln\tan\frac{x}{y}$; (6)$z=(1+xy)^{y}$; (7)$u=x^{\frac{y}{z}}$; (8)$u=arctan(x-y)^{z}$.
题目解答
答案
(5) $ z = \ln \tan \frac{x}{y} $
\[
\boxed{
\begin{aligned}
&\frac{\partial z}{\partial x} = \frac{2}{y} \csc \frac{2x}{y}, \\
&\frac{\partial z}{\partial y} = -\frac{2x}{y^2} \csc \frac{2x}{y}.
\end{aligned}
}
\]
(6) $ z = (1+xy)^y $
\[
\boxed{
\begin{aligned}
&\frac{\partial z}{\partial x} = y^2 (1+xy)^{y-1}, \\
&\frac{\partial z}{\partial y} = (1+xy)^y \left[ \ln (1+xy) + \frac{xy}{1+xy} \right].
\end{aligned}
}
\]
(7) $ u = x^{\frac{y}{z}} $
\[
\boxed{
\begin{aligned}
&\frac{\partial u}{\partial x} = \frac{y}{z} x^{\frac{y}{z} - 1}, \\
&\frac{\partial u}{\partial y} = x^{\frac{y}{z}} \cdot \frac{\ln x}{z}, \\
&\frac{\partial u}{\partial z} = -\frac{y \ln x}{z^2} x^{\frac{y}{z}}.
\end{aligned}
}
\]
(8) $ u = \arctan (x-y)^z $
\[
\boxed{
\begin{aligned}
&\frac{\partial u}{\partial x} = \frac{z (x-y)^{z-1}}{1 + (x-y)^{2z}}, \\
&\frac{\partial u}{\partial y} = -\frac{z (x-y)^{z-1}}{1 + (x-y)^{2z}}, \\
&\frac{\partial u}{\partial z} = \frac{(x-y)^z \ln (x-y)}{1 + (x-y)^{2z}}.
\end{aligned}
}
\]