= Solution
For fixed $X\in L(G)$, set $b(t)=\phi(e^{tX})$. The <homomorphism> property gives $b(s+t)=b(s)b(t)$ and $b(0)=I$. Its initial derivative is $L(\phi)(X)=D\phi_I(X)$. The <one-parameter subgroup> uniqueness theorem gives
$$
b(t)=e^{tL(\phi)(X)},\qquad\boxed{\phi(e^X)=e^{L(\phi)(X)}}.
$$
For the <Adjoint representation>, $\operatorname{Ad}_gY=gYg^{-1}$. Thus
$$
\boxed{\operatorname{ad}_XY=\left.\frac d{dt}(e^{tX}Ye^{-tX})\right|_{t=0}=XY-YX=[X,Y]}.
$$
The differential $L(\phi)$ is a real linear map of Lie algebras; the bracket identity follows by differentiating the compatibility of conjugation with $\phi$.
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