Lie derivative for a wedge product $\omega_{1}\wedge\omega_{2}$ 
Question: I have to prove that $$\mathcal L_X\omega_{1}\wedge\omega_{2}=(\mathcal L_X\omega_{1})\wedge\omega_{2}+\omega_{1}\wedge(\mathcal L_X\omega_{2})$$ 
  using the definition
  $$\mathcal L_X\omega=\frac{d}{dt}\bigg|_{t=0}\varphi_t^*\omega=\lim_{t\rightarrow 0} \frac{\varphi_t^*\omega-\omega}{t}$$

Attempt: I get
$$\mathcal L_X(\omega_{1}\wedge\omega_{2})=\frac{d}{dt}\bigg |_{t=0}\varphi_t^*(\omega_{1}\wedge\omega_{2})$$
The homework hint says: Use $\varphi_t^*(\omega_{1}\wedge\omega_{2})=(\varphi_t^*\omega_{1})\wedge(\varphi_t^*\omega_{2})$ and the result is immediate, but I can't see the next step and I figure that thes proof hidden a trick.
 A: Assuming $\omega \in \Omega^k(M)$ and $\eta \in \Omega^l(M)$. From where you stopped we have
$$\mathcal L_v (\omega \wedge \eta) = \frac{d}{dt}\bigg|_{t=0} \big((\varphi_t)^* \omega \wedge (\varphi_t)^*\eta\big) $$
Define $F: I \times I \to M$ by $F(t,s) = (\varphi_t)^* \omega \wedge (\varphi_s)^*\eta $ and consider $\delta : I \to I \times I$ the diagonal map given by $\delta (t) = (t,t)$. Thus by the chain rule 
$$\begin{aligned} \frac{d}{dt}\bigg|_{t=0} F(t,t) &= \frac{d}{dt}\bigg|_{t=0} (F \circ \delta )(t)  \\&= dF_{(0,0)} \cdot \delta' (0) \\&=dF_{(0,0)} \cdot (1,1) \\&= dF_{(0,0)} (1,0) + dF_{(0,0)}(0,1)\\&=  \frac{d}{dt}\bigg|_{t=0} F(t,0) +  \frac{d}{dt}\bigg|_{t=0} F(0,t)  \end{aligned}$$
Now remember that $\varphi_0 = id_M$ and  and it follows that 
$$\begin{aligned}\mathcal L_v (\omega \wedge \eta) &=  \frac{d}{dt}\bigg|_{t=0} \big((\varphi_t)^* \omega  \big)\wedge \eta +  \omega \wedge \frac{d}{dt}\bigg|_{t=0}  (\varphi_t)^*\eta\big)\\&=(\mathcal L _v \omega) \wedge \eta + \omega \wedge (\mathcal L_v \eta) \end{aligned} $$
since $(\varphi_0)^*\omega  = \omega $  and $(\varphi_0)^*\eta  = \eta $ don't depend on $t$. 
And we are done.  
A: Now use the product rule to calculate $\frac{d}{dt}\big\vert_{t=0}(\varphi_t^*\omega_1)\wedge(\varphi_t^*\omega_2)$.
