Find $u'(0)$ where $u(t) = f(2009t, t^{2009})$ given the differential of $f$ in $0$ I'm having trouble with this exercise, probably don't have enough theoretical info. How do I approach this?

Let $f : \mathbb{R}^2\rightarrow \mathbb{R}$ where $f$ - differentiable in $0$ and $df(0)(h) = 2h_{1}-7h_{2} $ for $h\in\mathbb{R}^2.$ Let $u : \mathbb{R} \rightarrow \mathbb{R} : u(t) = f(2009t, t^{2009})$. Find $u'(0)$ 

Thanks for your help!
 A: $$u(0) = f(0,0)$$ 
$$u'(t) = \frac{d u}{d t} = \frac{\partial f}{\partial x}\cdot\frac{d x}{dt} + \frac{\partial f}{\partial y}\cdot\frac{d y}{dt}$$
$$u'(0) = 2\cdot2009 -7 \cdot 2009(0)^{2008} = 4018$$
A: First of all, in response to nbubis's comment:  $df(0)h$ is presumably the differential of $f$ at zero expressed as an element of the cotangent space to $R^2$ at $0$, i.e. as a linear functional from $R^2$ (which is incidentally said cotangent space) to $R$, evaluated on the tangent vector $h$.  Thus if $h = (h_1, h_2)$ is a tangent vector to $R^2$ at $0$, the problem specifies that this linear functional is $2h_1 - 7h_2$.
In the light of the above, since in general $df = \frac{\partial f} {\partial x}dx + \frac{\partial f} {\partial y}dy$, where $x$, $y$ are the standard coordinates on $R^2$, we have $\frac{\partial f} {\partial x}(0) = 2$ and  $\frac{\partial f} {\partial y}(0) =-7$; we simply read these values off from the given expression $df(0)(h) = 2h_1 - 7h_2$.  It then follows from the chain rule that $u'(0) = {\frac{\partial f} {\partial x}}(0)[{\frac {d(2009t)}{dt}}]_{t = 0} + {\frac{\partial f} {\partial y}}(0)[{\frac{d(t^{2009})}{dt}}]_{t = 0}$, or by a simple calculation, $u'(0) = 2(2009) - 7(0) = 4018$.
