Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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 $t.$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.