접평면과 선형근사

By definition, the differential at $x=a$, $y=b$ is

$$f_x(a,b)\,\mathrm{d}x + f_y(a,b)\,\mathrm{d}y$$

so we have to determine the partial derivatives $f_x(a,b)$ and $f_y(a,b)$. We are told that

$$e^{f(x,y)} + y\,f(x,y) = x + y$$

for all $x$ and $y$. Differentiating this equation with respect to $x$ and with respect to $y$ gives, by the chain rule,

$$\begin{aligned} e^{f(x,y)}f_x(x,y) + y\,f_x(x,y) = 1 \\ e^{f(x,y)}f_y(x,y) + f(x,y) +y\,f_y(x,y) = 1 \end{aligned}$$

Solving the first equation for $f_x$ and the second for $f_y$ gives

$$\begin{aligned} f_x(x,y) &= \frac{1}{e^{f(x,y)}+y} \\ f_y(x,y) &= \frac{1-f(x,y)}{e^{f(x,y)}+y} \end{aligned}$$

So the differential at $x=a$, $y=b$ is

$$\frac{\mathrm{d}x}{e^{f(a,b)}+b} + \frac{1-f(a,b)}{e^{f(a,b)}+b} \,\mathrm{d}y$$

Since we can't solve explicitly for $f(a,b)$ for general $a$ and $b$. There's not much more that we can do with this.

(b) In particular, when $a=1$ and $b=0$, we have

$$e^{f(1,0)} + 0\,f(1,0) = 1 + 0 \implies e^{f(1,0)} = 1 \implies f(1,0)=0$$

and the linear approximation simpifies to

$$f\big(1+\mathrm{d}x\,,\,\mathrm{d}y\big) \approx f(1,0) + \frac{\mathrm{d}x}{e^{f(1,0)}+0} + \frac{1-f(1,0)}{e^{f(1,0)}+0} \,\mathrm{d}y = \mathrm{d}x +\mathrm{d}y$$

Choosing $\mathrm{d}x = -0.01$ and $\mathrm{d}y = 0.01$, we have

$$f\big(0.99\,,\,0.01\big) \approx -0.01 + 0.01 =0$$