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Let $M$ be a differentiable manifold and $f$ a differentiable mapping

$f : (-\epsilon, \epsilon) \times M \to M$, $f(t,x) = f_t(x)$

Furthermore let $\gamma: (-\epsilon,\epsilon) \to M$ be a differentiable curve through $p \in M$.

My question is:  Is there a formula for $\frac{d}{dt} \vert_{t=0} f_t(\gamma(t))$

Thanks in advance!

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Hint: In general, when you have a function $g:\mathbf R^n\times\mathbf R^p\to\mathbf R^m$ and functions $a:\mathbf R\to\mathbf R^n$ and $b:\mathbf R\to\mathbf R^p$, and you want to find the derivative of $\phi(t)=f(a(t),b(t))$, you have $ \frac{\mathrm d\phi}{\mathrm dt} = \frac{\partial f}{\partial x}(a(t),b(t))\cdot a'(t) + \frac{\partial f}{\partial y}(a(t),b(t))\cdot b'(t),$ where $\partial f/\partial x$ is the differential of the function $f_y:\mathbf R^n\to\mathbf R^m$ defined by $f_y(x)=f(x,y)$, and similarly for $\partial f/\partial y$.

Here you have a slightly more general case (with manifolds), but it's basically the same thing.

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    @thomas.en Assuming that $p=\gamma(0)$, what you wrote as $\nabla f$ is the gradient of $f$ *at $p$*.2012-12-03