Question on derivative

Question

I want to differentiate $H(p(t),q(t))=1 $ with respect to $t$, where $H:\mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R} $ is a convex function.

I think that it is:

$\displaystyle \frac{dp}{dt} \frac{\partial H(p,q)}{\partial p}+ \frac{dq}{dt} \frac{\partial H(p,q)}{\partial q}=0$

right?

Please, thank you.

$\displaystyle \frac{\partial H}{\partial t}=\frac{dp}{dt} \frac{\partial H(p,q)}{\partial p}+ \frac{dq}{dt} \frac{\partial H(p,q)}{\partial q}$ — M.H, May 21 '13 at 19:29
You wrote $\dots = 0$; the equals-zero isn't part of the derivative. (It's like saying "the derivative of $x^2$ is $2x=0$".) If you want to differentiate and then set the derivative equal to $0$ in a second step, that's fine, but it's important to recognize that those are indeed separate things to do. — Greg Martin, May 21 '13 at 19:55
so the defferential of $H(p(t),q(t))$ is $\displaystyle \frac{\partial H}{\partial t}=\frac{dp}{dt} \frac{\partial H(p,q)}{\partial p}+ \frac{dq}{dt} \frac{\partial H(p,q)}{\partial q}$, ans because $H(p(t),q(t))=1$ then $\displaystyle \frac{dp}{dt} \frac{\partial H(p,q)}{\partial p}+ \frac{dq}{dt} \frac{\partial H(p,q)}{\partial q}=0$ right ? — Vrouvrou, May 21 '13 at 19:59

score 1 · Accepted Answer · answered May 24 '13 at 00:50

If we want something we need a detailed notation flowery but enlightening.

First of all use the implicit differentiation theorem: \begin{align} \left.\frac{d\!}{dt}\!F\big( q(t), p(t)\big)\right|_{t=t_0} = & \left.\frac{d\!}{dt}\!F\big( q_0, p(t)\big)\right|_{t=t_0} + \left.\frac{d\!}{dt}\!F\big( q(t), p_0\big)\right|_{t=t_0} \end{align} for $p_0=p(t_0)$ and $q_0=q(t_0)$.
Second, use the chain rule

$\qquad\left.\dfrac{d\!}{dt}\!F\big( q_0, p(t)\big)\right|_{t=t_0}=\left.\dfrac{d}{dt}p(t)\right|_{t=t_0}\!\cdot\; \left.\dfrac{\partial}{\partial p}F(p,q)\right|_{(p,q)=(p_0,q_0)},$

$\qquad\left.\dfrac{d\!}{dt}\!F\big( q(t), p_0\big)\right|_{t=t_0}=\left.\dfrac{d}{dt}q(t)\right|_{t=t_0}\!\cdot\; \left.\dfrac{\partial}{\partial q}F(p,q)\right|_{(p,q)=(p_0,q_0)}$

to get \begin{align} \left.\frac{d\!}{dt}\!F\big( q(t), p(t)\big)\right|_{t=t_0} = & \left.\dfrac{d}{dt}p(t)\right|_{t=t_0}\!\cdot\; \left.\dfrac{\partial}{\partial p}F(p,q)\right|_{(p,q)=(p_0,q_0)}+ \\ + & \left.\dfrac{d}{dt}q(t)\right|_{t=t_0}\!\cdot\; \left.\dfrac{\partial}{\partial q}F(p,q)\right|_{(p,q)=(p_0,q_0)} \end{align}

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