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Following my previews question : From the comments to the answer I feel that I don't understand how to use the chain rule.

From what I understand the chain rule sais : $F(t)=f(x(t),y(t))\implies F'(t)=\langle(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}),(x'(t),y'(t))\rangle$. [with the right conditions]

Here $F(t)=f(x(t),y(t))$ where $x(t)=x+t,y(t)=y\implies x'(t)=1,y'(t)=0\implies{P_{x}(f)(v)=(\frac{d}{dt}(f(T_{\begin{pmatrix}t\\ 0 \end{pmatrix}}v))|_{t=0}=(\frac{\partial f}{\partial x}1+\frac{\partial f}{\partial y}0)(v)=\frac{\partial f}{\partial x}v}$.

So I concluded that the fact we evaluate at $t=0$ is irrelevant (that is evaluating at a different point will not change the answer), but I was told I am wrong (in the comments).

Am I using the chain rule wrong ?

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I think you're being led astray by having chose $x(t)$ and $y(t)$ such that $x'(t)$ and $y'(t)$ are both constant functions. Try with a less simple example -- such as $x(t)=t^2$, $y(t)=t^3$ -- and you will see that it does indeed matter where you evaluate the derivatives. – Henning Makholm Apr 21 '12 at 12:15
@HenningMakholm - I am referring to this case. is it dependent on t ? (I get that no...) – Belgi Apr 21 '12 at 13:22
up vote 1 down vote accepted

Taking the smooth maps $\gamma:t\in\mathbb{R}\to \gamma(t)\in\mathbb{R}^n$ and $f:x\in\mathbb{R}^n\to f(x)\in\mathbb{R},$ the chain rule for derivatives \sout{implies} states that: $$(f\circ\gamma)'(t)=\langle(\nabla f)_{(\gamma(t))},\gamma'(t)\rangle.$$ Where $(\nabla f)_{\gamma(t)}$ is $\nabla f=(\partial_{x_1} f,\ldots,\partial_{x_n}f):\mathbb{R}^n\to\mathbb{R}^n$ evaluated at $\gamma(t).$

So in your example even if $\gamma'(t)$ is independent of $t$ the complessive espression is dependent on the choice of $t.$

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How does it impky from my version of the cahin rule that you evaluate at $\gamma(t)$ ? – Belgi Apr 21 '12 at 13:05
You have incorrectly reported the chain rule, in my post I would point out for you its correct form. – Giuseppe Apr 21 '12 at 13:22

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