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Let $\phi$(x, y) be a function of variables x,y. I am trying to prove that if the total derivative d$\phi$ = $\phi_x dx + \phi_y dy$ is 0, then the partial derivatives $\phi_x , \phi_y$ are 0. The informal argument in the book goes like this

$\phi_x dx + \phi_y dy$ = 0

(1) $\rightarrow\phi_x + \phi_y \frac{dy}{dx}$ = 0

(2) $\rightarrow \phi_x = 0$

(3) $\rightarrow \phi$ is a function of only y

(4) $\rightarrow \phi_y = 0$

Any elucidating details would be appreciated, for I'm not sure if I follow any of this proof, especially how (1) implies (2).

fdzsfhaS
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1 Answers1

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The total derivative at $(a,b)$ of $\phi :\mathbb R^2\to \mathbb R$ is by definition a linear transformation $D\phi ((a,b)):\mathbb R^2\to \mathbb R$ that satisfies $\lim _{\vert h\vert \to 0}\frac{\vert \phi (a+h)-\phi (a )-{D\phi }(a)h\vert }{\vert h\vert }=0$.

If $\phi $ is differentiable at $a$=$(a,b)$ and if $h=(h_1,h_2),$ then it's easy to show that the partials esxist and that

$D\phi (a,b)(h_1,h_2)=\phi_x(a,b)\cdot h_1+\phi_y(a,b)\cdot h_2$.

If $D\phi (a,b)=0$ then, taking in turn $(h_1,h_2)=(1,0)$ and $(0,1)$ we get $\phi_x(a,b)=\phi_y(a,b)=0$.

Matematleta
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