# formula for derivative of differentiable mapping

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 $\frac{d}{dt} \vert_{t=0} f_t(\gamma(t))$ ?

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 $\partial f/\partial y$.
So I get $\frac{\partial f}{\partial x} (t, \gamma(t)) + \frac{\partial f}{\partial y} (t, \gamma(t)) \gamma'(t) \vert_{t=0}$, what is equal to $\nabla f (1, \gamma'(t))^{T} \vert_{t=0}$. How can I use the point p? –  thomas.en Dec 2 '12 at 11:17
@thomas.en Assuming that $p=\gamma(0)$, what you wrote as $\nabla f$ is the gradient of $f$ at $p$. –  jathd Dec 3 '12 at 0:09