If given an equation $$ z=e^{xy} $$ where $$ x = 1+st , y = s^2-t^2 $$
I found $$dz/dt , dz/ds$$ using the chain rule and got the following: $$ dz/ds = e^{xy}(yt+2xs)$$ $$dz/dt = e^{xy}(ys-2xt) $$ Suppose I am to provide a linear approximation of $z$ at $s = 1.1$ and $t = -0.2,$ how do I use the equations above to do so? Do I plug in the values of $s$ and $t$ in the equations of $dz$? Moreover, can I use $s = 1$ and $t = 0$ to find the linearization formula for $z$?
According to the Taylor formula, $$ z(s_0+\Delta s,t_0+\Delta t)=z(s_0,t_0)+\frac{\partial z}{\partial s}(s_0,t_0)\Delta s+ \frac{\partial z}{\partial t}(s_0,t_0)\Delta t+R_1(\Delta s,\Delta t). $$ Neglecting the remainder, we obtain (for small $\Delta s$, $\Delta t$) $$\tag{1} z(s_0+\Delta s,t_0+\Delta t)\approx z(s_0,t_0)+\frac{\partial z}{\partial s}(s_0,t_0)\Delta s+ \frac{\partial z}{\partial t}(s_0,t_0)\Delta t. $$ Since we have $s_0=1$, $t_0=0$, (1) transforms into $$\tag{2} z(1+\Delta s,0+\Delta t)\approx z(1,0)+\frac{\partial z}{\partial s}(1,0)\Delta s+ \frac{\partial z}{\partial t}(1,0)\Delta t. $$ Calculating the values at $(1,0)$, we get $$ x\Big|_{s=1,t=0}=1,\quad y\Big|_{s=1,t=0}=1,\quad z(1,0)= e^{1\cdot 1}=e, $$ $$ \frac{\partial z}{\partial s}(1,0)= e \cdot (0+2)=2e, \quad \frac{\partial z}{\partial t}(1,0)= e \cdot (1-0)=e. $$ Hence, (2) transforms into $$\tag{3} z(1+\Delta s,0+\Delta t)\approx e+2e\Delta s+e\Delta t. $$ For $s = 1.1$ and $t = -0.2$ we have $$\tag{4} z(1.1,-0.2)= z(1+0.1,0-0.2)\approx e+0.2e-0.2e=e. $$