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We know that the commutator of a lie algebra is defined as $$[M,N]=MN-NM.$$ I have seen On the relationship between the commutators of a Lie group and its Lie algebra. He has provided a proof for that. But I was reading the book Lie Algebras: Finite and Infinite Dimensional Lie Algebras and Applications in Physics Pt. 1. In that book it is mentioned that if $C(t)=A(t)B(t)A^{-1}(t)B^{-1}(t)$ then $$ \dot{C}(0)=MN-NM,\ \text{where } M=\dot{A}(0)\ \text{and } N=\dot{B}(0) . $$ But while computing the derivative of $C(t)$ I am getting \begin{align*} C^\prime(t) & =A(t)^\prime B(t)A^{-1}(t)B^{-1}(t)+A(t)B(t)^\prime A^{-1}(t)B^{-1}(t)+A(t)B(t)A^{-1}(t)^\prime B^{-1}(t)+A(t)B(t)A^{-1}(t)B^{-1}(t)^\prime\\ C^\prime(0) &= M+N-M-N=0 \end{align*} Will anybody please tell me where I am doing the mistake and how will I get the commutator relation. Second order approximation.

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    The name you want to google is "Baker-Campbell-Hausdorff Formula" – 2017-02-27

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You need to expand to second order in $t$. Try writing e.g. $A(t) = e^{tM} \approx I + tM + \frac{t^2}{2}M + ... $ and expand $ABA^{-1}B^{-1}$ to second order.

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    I have added the second order expansion, but still I am not getting. – 2017-02-26
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    Will you please give a more elaboration of your answer? – 2017-02-26
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    Expand $(I+tM+\frac{t^2}{2}M)(I+tN+\frac{t^2}{2}N)(I-tM+\frac{t^2}{2}M)(I-tN+\frac{t^2}{2}N)$ to second order in $t$. – 2017-02-26
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    The solution in your picture isn't correct. You can't drop the $\frac{t^2}2 M$-like terms in the brackets. – 2017-02-26
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    I am unable to get you. Will you please provide me the details? – 2017-02-26
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    Just do exactly what I said in my third comment. Treat it as a high school algebra exercise. – 2017-02-26