Given $f(x,y)=5+2x+4y+x^2+y^2+(x^2y^4)^\frac15$, show that it is differentiable at $(0,0)$.

Question

I was given the function:

$f(x,y)=5+2x+4y+x^2+y^2+(x^2y^4)^\frac15$

I need to show it is differentiable at $(0,0)$.

I started using the method of differentials and infinitesimal functions:

$\Delta f=f(0+\Delta x,0+\Delta y)-f(0,0)$

$\Delta f=2\Delta x + 4 \Delta y + (\Delta x)^2 + (\Delta y)^2 + ((\Delta x)^2(\Delta y)^4)^\frac15$

Now I cannot see a way how I can obtain infinitesimal functions* $\alpha$ and $\beta$ such that:

$\Delta f=2\Delta x + 4 \Delta y + \alpha\Delta x + \beta \Delta y$, where $\alpha,\beta=o(\rho)$

(Where $\rho^2 = x^2+y^2$, and $o$ is the Landau little-$o$ notation defined as $\alpha=o(\rho)\iff\lim \alpha/\rho=0$).

I managed to get 3 "infinitesimal functions": $\alpha=\Delta x$, $\beta=\Delta y$, and $\gamma=(\Delta x^2\Delta y^{-1})^\frac15$ but I am having trouble showing $\gamma=o(\rho)$ because frankly it doesn't seem to be.

Any hints? If this is not possible, are there any other methods that one would recommend?

*My prof defines infinitesimal functions as: $\alpha(x)$ is infinitesimal at $a$ if $\lim_{x\rightarrow a}\alpha(x)=0$

Answer 1

You seem to have the wrong definition: You need only $\Delta f=2\Delta x + 4 \Delta y + \alpha\Delta x + \beta \Delta y$, where $\alpha,\beta=o(1).$ An equivalent form is $\Delta f=2\Delta x + 4 \Delta y + o(\rho),$ which I think is more standard. If you use the latter form, you need only verify that

$$ (\Delta x)^2 + (\Delta y)^2 + ((\Delta x)^2(\Delta y)^4)^\frac15=o(\rho).$$

Answer 2

We know that a function f(x,y) is differentiable at a point (x,y) if it's left hand derivative is equal to it's right hand derivative (please read the differentiation chapter taught in your respective school/college).

That is, $$lim_{h\to 0}\frac{f(x+h,y+h)-f(x,y)}{h}=lim_{h\to 0}\frac{f(x-h,y-h)-f(x,y)}{-h}$$

Where h is a positive number that lies between 0 and 1 and can be approximately equated to 0. I leave the calculation part up to you(Just put 0 in place of x and y and apply the limits) You will get a value 6 in the left hand derivative and the same value for the right hand derivative which proves that the function is indeed differentiable at (0,0). I will be very happy to post the calculation if you are not getting it. Hope this solves your query.

Answer 3

Another approach using the definition of differentiability:

To use the horrible limit

$$\lim_{(x,y)\to(0,0)}{\dfrac{f(x,y)-[f(0,0)+f'_x(0,0)(x-0)+f'_y(0,0)(y-0)]}{\sqrt{{(x-0)}^2+{(y-0)}^2}}}\tag 1$$

we need first to calculate the partial derivatives at $(0,0)$ using the fact that $f$ is $C^1(\text{Dom}(f))$. Let's begin.

Remember that $f(x,y)=5+2x+4y+x^2+y^2+\sqrt[5]{x^2y^4}$.

Finding the domain of the function

$\text{Dom}(f)=\mathbb R^2$ because is a sum of continuous.

Finding partial derivatives

$$\begin{array}{llc} f'((0,0);(a,b))&&= \\ &\displaystyle\lim_{h\to 0}{\dfrac{f(ah,bh)-f(0,0)}{h}}&= \\ &\displaystyle\lim_{h\to 0}{\dfrac{\left(5+2ah+4bh+a^2h^2+b^2h^2+\sqrt[5]{a^2h^2b^4h^4}\right)-5}{h}}&= \\ &\displaystyle\lim_{h\to 0}{\dfrac{h\left(2a+4b+a^2h+b^2h+\frac{\sqrt[5]{a^2b^4h^6}}{h}\right)}{h}}&= \\ &\displaystyle\lim_{h\to 0}{\left(2a+4b+a^2h+b^2h+\sqrt[5]{a^2b^4h}\right)}&= \\ &2a+4b. \end{array}$$

Thus the directional derivatives are continuous for all $(x,y)\in\mathbb R^2$, in particular for $(x,y)=(0,0)$, thus $f(x,y)\in C^1\left(\mathbb R^2\right)$ so

$$\begin{matrix} f'((0,0);(1,0))&=&f'_x(0,0)&=&2\cdot 1+4\cdot 0&=&2, \\ f'((0,0);(0,1))&=&f'_y(0,0)&=&2\cdot 0+4\cdot 1&=&4. \end{matrix}$$

Proving the differentiability

From $(1)$ $$\begin{array}{lcc} \displaystyle\lim_{(x,y)\to(0,0)}{\dfrac{5+2x+4y+x^2+y^2+\sqrt[5]{x^2y^4}-[5+2(x-0)+4(y-0)]}{\sqrt{x^2+y^2}}} &=& \\ \displaystyle\lim_{(x,y)\to(0,0)}{\dfrac{x^2+y^2+\sqrt[5]{x^2y^4}}{\sqrt{x^2+y^2}}}&\underbrace =_{(x,\,y)\;=\;(\rho\cos\theta,\,\rho\sin\theta)}&\\ \displaystyle{\lim_{\rho\to 0^+}}_{\theta\in[0,2\pi)}{\dfrac{\rho^2\cos^2\theta+\rho^2\sin^2\theta+\sqrt[5]{\rho^2\cos^2\theta\cdot\rho^4\sin^4\theta}}{\sqrt{\rho^2\cos^2\theta+\rho^2\sin^2\theta}}}&=&\\ \displaystyle{\lim_{\rho\to 0^+}}_{\theta\in[0,2\pi)}{\dfrac{\rho^2+\rho\sqrt[5]{p\cos^2\theta\cdot\sin^4\theta}}{\rho}}&=&\\ \displaystyle{\lim_{\rho\to 0^+}}_{\theta\in[0,2\pi)}{\left(\rho+\sqrt[5]{p\cos^2\theta\cdot\sin^4\theta}\right)}&=&\\ 0,&& \end{array}$$ hence $$\boxed{f(x,y)\quad\text{is differentiable at}\quad(0,0)}.$$