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I'm trying to show the following but I'm not very sure how to proceed. Could someone please explain to me how to approach and solve the following question?

Let $f : \Bbb R^2 → \Bbb R$ be differentiable such that $f(x, 2x) = 1$ and $f (−x, x) = 1 \; ∀x ∈ \Bbb R$. Using the Chain Rule, show that $∇f (0, 0) = (0, 0).$

nTuply
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1 Answers1

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Let $\varphi_{1}: x \mapsto (x, 2x)$ and let $$g_{1}: x \overset{\varphi_{1}}{\mapsto} (x, 2x) =: (u, v) \overset{f}{\mapsto} f(u, v) = 1$$ on $\mathbb{R}$. Then $$g'_{1}(x) = 0 = \big( D_{1}f(x, 2x), D_{2}f(x, 2x) \big)\cdot (1, 2) = D_{1}f(x, 2x) + 2D_{2}f(x, 2x)$$ for all $x \in \mathbb{R}.$

Let $\varphi_{2}: x \mapsto (-x, x)$ and let $$g_{2}: x \overset{\varphi_{2}}{\mapsto} (-x, x) =: (u, v) \overset{f}{\mapsto} f(u, v) = 1$$ on $\mathbb{R}$. Then $$g'_{2}(x) = 0 = \big( D_{1}f(-x, x), D_{2}f(-x, x) \big)\cdot (-1, 1) = -D_{1}f(-x, x) + D_{2}f(-x, x)$$ for all $x \in \mathbb{R}.$ In sum, we see that $$\nabla f(0, 0) = (0, 0).$$

Note: The $(u, v)$ seems redundant, but I just want to show you in slow motion.

Yes
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