Covariant derivative identity I am trying to prove the following identity for contravariant vectors $X$ and $Y$ (this appears in exercise 6.7 of D'Inverno):
$\nabla_{X}(fY) = (Xf)Y + f \nabla_X Y$.
I have a way of proving it but it assumes linearity of the function $f$, whereas the proof is supposed to hold if $f$ is any smooth function. Here's my proof:
\begin{eqnarray*}
\nabla_{X}(fY) &=& X^b \nabla_b(fY^a)\\
 &=& X^b(\partial_b (fY^a) + fY^c\Gamma^{a}_{cb})\\ 
&=& X^b(\partial_bf(Y^a) + f\partial_b Y^a + fY^c \Gamma^{a}_{cb})\\
&=&(X^b \partial_b f)Y^a + X^bf(\partial_bY^a + Y^c\Gamma^{a}_{cb})\\
&=& (Xf)Y + f(X^b\nabla_bY^a)\\
&=& (Xf)Y + f \nabla_X Y
\end{eqnarray*}
Is there a better way to prove this without assuming linearity of $f$? I have searched around for a proof of this to no avail.
 A: your formula or at least notation is weird. 
let me define the covariant derivative with the parallel transport  $T_a^b : T_aM \to T_b M$  instead of the Christoffel symbols $\Gamma_{cb}^a$. 
for a point $p \in M$ and $v \in T_pM$ :
$$ \nabla_v Y (p) = \frac{\partial T_{p+hv}^p Y(p+hv)}{\left.\partial h \right|_{h=0}}= \frac{\partial T_{p+hv}^p }{\left.\partial h \right|_{h=0}} Y(p) + \frac{\partial Y(p+hv)}{\left.\partial h \right|_{h=0}}$$
so $$\begin{eqnarray}\nabla_v (f Y) (p)&=& \frac{\partial T_{p+hv}^p }{\left.\partial h \right|_{h=0}} (fY)(p) + \frac{\partial f(p+hv)Y(p+hv)}{\left.\partial h \right|_{h=0}}\\ &=& f(p) \frac{\partial T_{p+hv}^p }{\left.\partial h \right|_{h=0}} Y(p)  + f(p) \frac{\partial Y(p+hv)}{\left.\partial h \right|_{h=0}} + Y(p) \frac{\partial f(p+hv)}{\left.\partial h \right|_{h=0}}\\&=& f(p) \nabla_v Y (p) + Y(p)\partial_v f(p)\end{eqnarray}$$
and because $(\nabla_X Y)(p) = \nabla_{X(p)} Y (p)$ we get :
$$\nabla_X (fY) = f \nabla_X Y + Y\partial_X f$$
(which is what we expect)
A: I believe I was thinking of the objects after $f$ as being arguments of $f$ rather than being multiplied by $f$. This was assumed in my argument, so the linearity assumption was not used.
