Differentiability of the function $ f(x, y) = | e ^x-y | (e^x-1) $

Question

Prove that the function $ f: \mathbb R^2 \to \mathbb R $ defined by the formula $ f(x, y) = | e ^x-y | (e^x-1) $ is differentiable at $ (a, b ) \in \mathbb R ^ 2 $ if and only if $ e ^ a \neq b $ or $ a = 0, b = 1 $.

I know that function is differentiable at $(x_0,y_0)$ if exist $Df(x_0,y_0)$ such that: $$\lim_{(h,k) \to (x_0,y_0)} \frac{f(x_0+h, y_0+k)-f(x_0,y_0)-h\cdot \frac{\partial f}{\partial x}(x_0,y_0)-k \cdot \frac{\partial f}{\partial y}(x_0,y_0)}{\sqrt{h^2+k^2}}=0$$

However, in my opinion, the formula of $f$ is too complicated to use this fact and I think that exist better solution.

Answer 1

We divide the plane in three sets. The region $S_{1}=\left\{(x,y):e^{x}-y>0 \right\}$ and the region $S_{2}=\left\{(x,y):e^{x}-y<0 \right\}$ and the set $S_{3}=\left\{(x,y):e^{x}=y \right\}$. First consider an $(a,b)\in S_{1}$. Then there is a neighborhood $U$ of $(a,b) \subseteq S_{1}$.

We will consider Gateaux derivative in direction $(h_{1},h_{2})$ to make things simpler and then we will invoke the following standard result by Drabek-Milota:

If $f:X\to Y$ is a map of linear normed spaces such that in a certain open neighbourhood $O_{x_0}$ of $x_0\in X$ the Gateaux differential $\delta f(x)$ exists and is linear and continuous for all $x\in O_{x_0}$ and the map $x\mapsto \delta f(x)$ from $X$ into $L(X,Y)$ is continuous at $x_0$, then f is Frechet differentiable at $x_0$.

The function in $U$ is $f(x,y)=(e^{x}-y)(e^{x}-1)$. So

$$ \begin{align} &\lim_{\epsilon \to 0}\dfrac{(e^{a+\epsilon\,h_{1}}-b-\epsilon\,h_{2})(e^{a+\epsilon\,h_{1}}-1)}{\epsilon}-\dfrac{(e^{a}-b)(e^{a}-1)}{\epsilon}\\ =&\lim_{\epsilon \to 0}\dfrac{e^{2a}(e^{2\epsilon\,h_{1}}-1)}{\epsilon}-e^{a}\dfrac{e^{\epsilon\,h_{1}}-1}{\epsilon}+be^{a}\dfrac{(1-e^{\epsilon\,h_{1}})}{\epsilon}+h_{2}\\ =&2h_{1}e^{2a}-h_{1}e^{a}-be^{b}h_{1}+h_{2} \end{align}$$ which is linear in $(h_{1},h_{2})$ and continuous.

And also $(x,y)\to\,\delta\,f(x,y)$ is continuous at $(a,b)$. Therefore $f(x,y)$ is Frechet differentiable at $(a,b)$.

Likewise for $(a,b)\in\,S_{2}$. Now assume $(a,b)\in S_{3}$ and $(a,b)\,\neq\,(0,1)$. Then the limit $$\lim_{\epsilon \to 0}\dfrac{e^{2a+2\epsilon\,h_{1}}-e^{a+\epsilon\,h_{1}}-be^{a+\epsilon\,h_{1}}+b}{\epsilon}-h_{2}e^{a}+h_{2}$$ does not exist unless $a=0$ and $b=1$. Thus we have proved the statement.

Answer 2

First of all, we will prove that

$\left|e^h-1\right|<e\left|h\right|\;,\quad\forall\,h\in\left]-1,1\right[\setminus\left\{0\right\}\,.\quad\color{blue}{(1)}$

Let $\,h\,$ be any real number in $\,\left]0,1\right[\,.$

Let $\,g_1:\left[0,h\right]\to\mathbb R\,$ be the function defined by

$g_1(x)=e^x-1\;,\quad\forall\,x\in\left[0,h\right]\,.$

Since the function $\,g_1\,$ is continuous on $\,\left[0,h\right]\,$ and differentiable on $\,\left]0,h\right[\,,$ we can apply the Mean Value Theorem, consequently there exists $\,c\in\left]0,h\right[\subseteq\left]0,1\right[\,$ such that $\,g_1(h)-g_1(0)=g_1’(c)\,h\;,\;$ hence ,

$\left|e^h-1\right|=e^h-1=e^ch<eh=e\left|h\right|\,.$

So far we have proved that

$\left|e^h-1\right|<e\left|h\right|\;,\quad\forall\,h\in\left]0,1\right[\,.$

Let $\,h\,$ be any real number in $\,\left]-1,0\right[\,.$

Let $\,g_2:\left[h,0\right]\to\mathbb R\,$ be the function defined by

$g_2(x)=e^x-1\;,\quad\forall\,x\in\left[h,0\right]\,.$

Since the function $\,g_2\,$ is continuous on $\,\left[h,0\right]\,$ and differentiable on $\,\left]h,0\right[\,,$ we can apply the Mean Value Theorem, consequently there exists $\,d\in\left]h,0\right[\subseteq\left]-1,0\right[\,$ such that $\,g_2(0)-g_2(h)=g_2’(d)\cdot(-h)\;,\;$ hence ,

$\left|e^h-1\right|=1-e^h=e^d(-h)<-h<-eh=e\left|h\right|\,.$

So we have also proved that

$\left|e^h-1\right|<e\left|h\right|\;,\quad\forall\,h\in\left]-1,0\right[\,.$

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As a consequence of $\,(1)\,,\,$ it results that

$\left|e^h-1\right|\leqslant e\left|h\right|\;,\quad\forall\,h\in\left]-1,1\right[\,.\quad\color{blue}{(2)}$

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Prove that the function $\,f:\mathbb R^2\to\mathbb R $ defined by the formula $\,f(x,y)=\left|e^x-y\right|\left(e^x-1\right)\,$ is differentiable at $\,(a,b)\in\mathbb R^2\,$ if and only if $\;e^a\!\neq\!b\,$ or $\,a\!=\!0\,,\,b\!=\!1$.

Proof :

For any $\,(a,b)\in\mathbb R^2\,$ such that $\,e^a\neq b\,,\,$ there exist both partial derivatives of the function $\,f(x,y)\,$ at the point $\,(a,b)\,$ and it results that

$\dfrac{\partial f}{\partial x}\big(a,b\big)=e^a\big|e^a-b\big|\left[\dfrac{e^a-1}{e^a-b}+1\right]\,,$

$\dfrac{\partial f}{\partial y}\big(a,b\big)=-\dfrac{\left|e^a-b\right|}{e^a-b}\big(e^a-1\big)\,.$

Since both partial derivatives $\,\dfrac{\partial f}{\partial x}\,$ and $\,\dfrac{\partial f}{\partial y}\,$ are continuous functions on the open region $A\!=\!\!\big\{(a,b)\,\big|\,(a,b)\!\in\!\mathbb R^2\!\land e^a\!\neq\!b\big\},$ the function $\,f\,$ is differentiable on $\,A\,.$

Now, for any $\,(a,b)\in\mathbb R^2\,$ such that $\,e^a=b\,,\,$ we will calculate the two following limits :

$\lim_\limits{h\to0^+}\dfrac{f(a+h,b)-f(a,b)}h\;\;\;;\quad\lim_\limits{h\to0^-}\dfrac{f(a+h,b)-f(a,b)}h\;\;.$

First we will calculate the right limit :

$\lim_\limits{h\to0^+}\dfrac{f(a+h,b)-f(a,b)}h=\lim_\limits{h\to0^+}\dfrac{\left|e^{a+h}-b\right|\left(e^{a+h}-1\right)}h=$

$=\!\lim_\limits{h\to0^+}\dfrac{\left|e^{a+h}-e^a\right|\left(e^{a+h}-1\right)}h\!=\!\lim_\limits{h\to0^+}\dfrac{e^a\left|e^h-1\right|\left(e^{a+h}-1\right)}h\!=$

$=\lim_\limits{h\to0^+}\dfrac{e^a\left(e^h-1\right)\left(e^{a+h}-1\right)}h=e^a\big(e^a-1\big)\,.$

Now we will calculate the left limit :

$\lim_\limits{h\to0^-}\dfrac{f(a+h,b)-f(a,b)}h=\lim_\limits{h\to0^-}\dfrac{\left|e^{a+h}-b\right|\left(e^{a+h}-1\right)}h=$

$\!=\!\lim_\limits{h\to0^-}\dfrac{\left|e^{a+h}-e^a\right|\left(e^{a+h}-1\right)}h\!=\!\lim_\limits{h\to0^-}\dfrac{e^a\left|e^h-1\right|\left(e^{a+h}-1\right)}h\!=$

$=\lim_\limits{h\to0^-}\dfrac{e^a\left(1-e^h\right)\left(e^{a+h}-1\right)}h=-e^a\big(e^a-1\big)\,.$

Consequently, the limit

$\lim_\limits{h\to0}\dfrac{f(a+h,b)-f(a,b)}h$

exists if and only if

$e^a\big(e^a-1\big)=-e^a\big(e^a-1\big)\;,$

$2e^a\big(e^a-1\big)=0\;,$

$a=0\,.$

Moreover , if $\,a=0\,,\,$ then $\;\lim_\limits{h\to0}\dfrac{f(a+h,b)-f(a,b)}h=0\,.$

It means that, for any $\,(a,b)\in\mathbb R^2\,$ such that $\,e^a=b\,,\,$ the partial derivative $\,\dfrac{\partial f}{\partial x}\,$ at the point $\,(a,b)\,$ exists if and only if $\;a=0\,,\,b=1\,.$

Moreover, it results that $\;\dfrac{\partial f}{\partial x}\big(0,1\big)=0\,.$

Now we will calculate the partial derivative $\;\dfrac{\partial f}{\partial y}\big(0,1\big)\,.$

$\dfrac{\partial f}{\partial y}\big(0,1\big)=\lim_\limits{k\to0}\dfrac{f(0,1+k)-f(0,1)}k=\lim_\limits{k\to0}\dfrac{\left|-k\right|\cdot0}k=0\,.$

The function $\,f\,$ cannot be differentiable at any $\,(a,b)\in\mathbb R^2\,$ such that $\,e^a=b\,$ and $\,a\neq0\,,\,$ indeed at those points there does not exist the partial derivative of $\,f\,$ with respect to $\,x\,.$

Finally we will prove that the function $\,f(x,y)\,$ is differentiable at the point $\,(0,1)\,.$

For any $\,h\in\left]-1,1\right[\,$ and for any $\,k\in\mathbb R\,$ such that $\,(h,k)\neq(0,0)\,,\,$ it results that

$\left|\dfrac{f(h,1+k)-f(0,1)-h\cdot\dfrac{\partial f}{\partial x}\big(0,1\big)-k\cdot\dfrac{\partial f}{\partial y}\big(0,1\big)}{\sqrt{h^2+k^2}}\,\right|=$

$=\dfrac{\left|e^h-1-k\right|\left|e^h-1\right|}{\sqrt{h^2+k^2}}\leqslant\dfrac{\big(\left|e^h-1\right|+\left|k\right|\big)\left|e^h-1\right|}{\sqrt{h^2+k^2}}\leqslant$

$\underset{\overbrace{\text{by using the inequality }(2)}}{\leqslant} \!\!\!\!\!\!\dfrac{\big(e\left|h\right|+\left|k\right|\big)e\left|h\right|}{\sqrt{h^2+k^2}}\leqslant\dfrac{\big(e\left|h\right|+e\left|k\right|\big)e\left|h\right|}{\sqrt{h^2+k^2}}=$

$=\!\dfrac{e^2\big(\!\left|h\right|\!+\!\left|k\right|\!\big)\left|h\right|}{\sqrt{h^2+k^2}}\!\leqslant\!\dfrac{e^2\big(\sqrt{h^2\!+\!k^2}+\sqrt{h^2\!+\!k^2}\big)\sqrt{h^2\!+\!k^2}}{\sqrt{h^2+k^2}}\!=$

$=2e^2\sqrt{h^2+k^2}\,.$

Since

$\left|\dfrac{f(h,\!1\!+\!k)\!-\!f(0,\!1)\!-\!h\dfrac{\partial f}{\partial x}\!\big(0,\!1\big)\!-k\!\dfrac{\partial f}{\partial y}\!\big(0,\!1\big)}{\sqrt{h^2+k^2}}\,\right|\leqslant2e^2\sqrt{h^2\!+\!k^2}$

for any $\,h\!\in\!\left]-1,1\right[\,$ and for any $\,k\!\in\!\mathbb R\,$ such that $\,(h,k)\!\neq\!(0,0)$

and since $\,\lim_\limits{(h,k)\to(0,0)}2e^2\sqrt{h^2+k^2}=0\,,\,$ it follows that

$\lim_\limits{(h,k)\to(0,0)}\!\!\!\!\dfrac{f(h,1\!+\!k)-f(0,1)-h\dfrac{\partial f}{\partial x}\big(0,1\big)-k\dfrac{\partial f}{\partial y}\big(0,1\big)}{\sqrt{h^2+k^2}}=0\;,$

consequently the function $f$ is differentiable at the point $(0,1).$

Hence the function $\,f\,$ is differentiable on the set $\,D\!=\!A\cup\big\{(0,1)\big\}\!=\!\big\{(a,b)\,\big|\,(a,b)\in\mathbb R^2\!\land e^a\!\neq\!b\big\}\cup\big\{(0,1)\big\}\,,$

but it is not differentiable on the set

$\mathbb R^2\setminus D=\big\{(a,b)\,\big|\,(a,b)\in\mathbb R^2\,\land\,e^a=b\,\land\,a\neq0\big\}\,.$

Answer 3

$f$ is obviously $C^\infty$ on the open set $\{(x,y)\in\mathbb R^2\mid y\ne e^x\}$.

Assume now $y_0=e^{x_0}$. Then, $$f(x_0+h,y_0+k)=y_0\left|e^h-1-\frac k{y_0}\right|(y_0e^h-1)=y_0\left|h-\frac k{y_0}+o(h)\right|(y_0-1+y_0h+o(h)).$$ As a particular case, $f(h,1+k)=o(\|(h,k)\|)$ hence $Df(0,1)=0$. But in the other cases ($y_0\ne1$), $f$ is not differentiable at $(x_0,y_0)$, since e.g. $$f(x_0,y_0+k)=|k|(y_0-1)\ne Ck+o(k)\quad(\forall C\in\mathbb R).$$