Answer-verification: Show that $f(x,y)=1+2x+3y$ for all $(x,y)\in \Bbb R^2$

Question

Define the function $f: \Bbb R^2\to \Bbb R$ has first order partial derivatives and that $f(0,0)=1$ While $\frac{\partial f}{\partial x}(x,y)=2$ and $\frac{\partial f}{\partial y}(x,y)=3$ for all $(x,y)\in \Bbb R^2$

Prove that $f(x,y)=1+2x+3y$ for all $(x,y)\in \Bbb R^2$

$\bf{solution:}$

Let $\frac{\partial f}{\partial x}(x,y)=g(x,y)$

Then $\int g(x,y)dx=2x +c_1$ for $c_1 $ is constant

Similarly, let $\frac{\partial f}{\partial y}=h(x,y)$

Then $\int h(x,y)=2y+c_2$ for $c_2$ constant

Since $f(0,0)=1$ is constant so, $c:=(c_1,c_2)=1$

Thus, $f(x,y)=1+2x+3y$ for all $(x,y)\in \Bbb R^2$

Is this answer true? If there exist mistakes, please can somebody correct this?

How? Explain more please.. Are the other parts correct? @AmireBendjeddou — Mathlover4, Jul 11 '13 at 19:32
I'm not sure about it. I deleted my comment , there will be more precise answers. — sigmatau, Jul 11 '13 at 19:35
When you integrate w.r.t. $x$, you should add an undetermined function of $y$ rather than a constant. Then differentiate that result w.r.t. y and it should work. — coffeemath, Jul 11 '13 at 19:43

Git Gud · Accepted Answer · 2013-07-11T19:57:32.740

4

Then $\int g(x,y)dx=2x +c_1$ for $c_1 $ is constant

This isn't correct. For instance take $\alpha (x,y)=2x+2y^3+\log (|y|+1)$. Then $\dfrac{\partial \alpha }{\partial x}(x,y)=2$ and $\alpha (x,y) \neq 2x$.

Correct would be: then $\int g(x,y)dx=2x+\varphi (y)=f(x,y)$, for some differentiable function $\varphi$.

Now you differentiate the above with respect to $y$ to get $$3=\frac{\partial f}{\partial y}(x,y)=\varphi '(y).$$

Therefore $\varphi (y)=3y+C$, for some constant $C\in \Bbb R$.

Now replacing $\varphi$ by what was just gotten it follows that $f(x,y)=2x+3y+C$.

Edit: You're also given that $f(0,0)=1$, so $2\cdot 0+3\cdot 0+C=1$ and it follows that $C=1$.

edited Jul 11 '13 at 19:57

answered Jul 11 '13 at 19:44

Git Gud

31,706

Should it be $d$ or $\partial$? – preferred_anon Jul 11 '13 at 19:47
@DanielLittlewood Is there some sort of convention on the meaning of each? I use them indistinctively. Is $\partial$ used for multivariable functions? – Git Gud Jul 11 '13 at 19:48
As far as I know, $d$ is used for ordinary derivatives and $\partial$ is for partial derivatives. With ordinary derivatives, the chain rule applies, and one would differentiate $\alpha$ implicitly. – preferred_anon Jul 11 '13 at 19:56
@DanielLittlewood I didn't know about that. I'll replace the $d$'s. – Git Gud Jul 11 '13 at 19:58
Okay. Okay I dont know rules @GitGud – Mathlover4 Jul 11 '13 at 20:09
@Rsm1 It's not a rule, just a suggestion of mine to make things look cleaner. – Git Gud Jul 11 '13 at 20:29

Answer-verification: Show that $f(x,y)=1+2x+3y$ for all $(x,y)\in \Bbb R^2$

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