I am learning some Differential Geometry on my own in preparation for a course I'm starting in October, and one of the exercises in the notes I'm using is to check that the Lie Derivative satisfies the Leibniz rule for tensors, or
$$ \mathcal{L}_{V}[A_1\otimes A_2 ]= \mathcal{L}_{V}[A_1] \otimes A_2 + A_1 \otimes \mathcal{L}_{V}[A_2]. $$
I have the definition of the Lie Derivative,
$$ \mathcal{L}_{V}[A]|_{p}=\lim_{\epsilon\rightarrow 0} \bigg{(}\frac{\sigma_{V}(+\epsilon)^{\ast}A|_{p'}-A|_p}{\epsilon}\bigg{)} $$
My first attempt was to plug the expression straight into the definition, resulting in
$$ \mathcal{L}_{V}[A_1\otimes A_2 ]|_{(p_1,p_2)}=\lim_{\epsilon\rightarrow 0} \bigg{(}\frac{\sigma_{V}(+\epsilon)^{\ast} (A_1 \otimes A_2)|_{(p_{1}',p_{2}')}-(A_1 \otimes A_2)|_{(p_{1},p_{2})}}{\epsilon}\bigg{)}, $$
but I'm at a bit of a loss as to what to do to from here on to be honest. My next step was to separate out the action of the pullback of the flow, ($\sigma_{V}(+\epsilon)^{\ast}$ onto the tensor product of the two tensors, resulting in
$$ \mathcal{L}_{V}[A_1\otimes A_2 ]|_{(p_1,p_2)}=\lim_{\epsilon\rightarrow 0} \bigg{(}\frac{(\sigma_{V}(+\epsilon)^{\ast} A_1)|_{p_1'} \otimes (\sigma_{V}(+\epsilon)^{\ast} A_2)|_{p_2'}-(A_1 \otimes A_2)|_{(p_{1},p_{2})}}{\epsilon}\bigg{)}, $$
but I'm essentially stumped. I'm new to Differential Geometry as a subject and am self studying so baby steps in explanations would be greatly appreciated.
Many thanks.