Calculate total derivative directly.

Question

Calculate directly (not via partial differentiation) the total derivative of the function $f(x_1,x_2)=x_1^2-10x_2.$ You may wish to use the fact that $\sqrt{x^2+y^2}\geq\frac{x+y}{2}.$ $$$$ For the solution I used the following formula: $f(x+y)=f(x)+df_x(y)+r(y), \text{where $\displaystyle\lim_{y\to0}\frac{\ r(y)}{||y||}=0$$$}$

So, $f(x+y)=(x_1+y_1)^2-10(x_2+y_2)\\ \hspace{1.6cm}=x_1^2+2x_1y_1+y_1^2-10x_2-10y_2 \\ \hspace{1.6cm}=f(x_1,x_2)+2x_1y_1+f(y_1,y_2)\\ \hspace{1.6cm}\text{here $r(y)=f(y_1,y_2)$, and I know that total derivative $(dfx(y))=2x_1y_1$ but I}\\ \text{don't know how to show that}\displaystyle\lim_{y\to0}\frac{\ r(y)}{||y||}=\displaystyle\lim_{y\to0}\frac{f(y_1, y_2)}{||y||}=0.\\ \text{Any suggestion would be appreciated!}$

Answer 1

The derivative is linear in the difference between the vectors $(x_1,y_1),(x_2,y_2)$, so it is probably better to write $$f(x+\Delta x,y+\Delta y)-f(x,y)=(x+\Delta x)^2-10(y+\Delta y)-x^2+10y$$ and what you want to do with this expression is to identify "what is linear in $\Delta x,\Delta y$ and prove that the rest is $o(\|(\Delta x,\Delta y)\|)$. So you expand and get $$f(x+\Delta x,y+\Delta y)-f(x,y)=\Delta x^2+(2x,-10)\cdot(\Delta x,\Delta y)$$

(Here $v\cdot u$ denotes the dot product ). Since $\frac{\Delta x^2}{\|(\Delta x,\Delta y)\|}\to 0$ as $(\Delta x,\Delta y)\to (0,0)$, we see that the linear map defined by dot-product with the vector $(2x,-10)$ is the total derivative of $f$ at the point $(x,y)$, as you would expect.

Answer 2

Derivative of a function $f\colon \Bbb R^2\to \Bbb R$ in $(x_1,x_2)$ is defined as a linear operator $A\colon\Bbb R^2\to\Bbb R$ such that $$ \frac{f((x_1,x_2)+(s,t))-f(x_1,x_2)-A(s,t)}{\|(s,t)\|}\to 0.$$ Since the operator $A$ is linear, it is of the form $A(s,t)=as+br$, where $a,b\in\Bbb R$. The operator $A$ of the form is a derivative of $f$ in $(x_1,x_2)$ iff
$$0\leftarrow \frac{f(x_1+s,x_2+t)-f(x_1,x_2)-as-bt}{\|(s,t)\|} = \frac{(x_1+s)^2 -10(x_2+t)-x_1^2+10x_2-as-bt}{\|(s,t)\|} = \frac{(2x_1-a)s+s^2 -(10+b)t}{\|(s,t)\|}=(*). $$ We have $$|(*)| \leq |2x_1-a+s|\cdot \frac {|s|}{\|(s,t)\|}+|10+b|\cdot \frac {|t|}{\|(s,t)\|} \leq |2x_1-a+s|+|10+b|. $$ We see that if $a=2x_1$ and $b=-10$ then $|(*)|\leq |s|\to 0$ if $(s,t)\to (0,0)$, that is the desired convergence hold. Therefore the operator $A(s,t)=2x_1s-10t$ is the derivative of $f$ in $(x_1,y_1)$.

Remark: Observe that the convergence in the product of spaces is a convergence componentwise, that is $(s,t)\to (0,0) \iff s,t\to 0$.