2-metric $\gamma_{AB}$ induced on the world sheet by the spacetime metric $g_{\mu\nu}$ is $$\gamma_{AB}=g_{\mu\nu}X^{\mu},_A X^{\nu},_B$$
$$\gamma^{AB},_B=-\gamma^{AC}\gamma^{BD}\gamma_{CD},_B$$
How is the derivative of the induced metric explicitly obtained? I guess it is derived from $\gamma^{AB}\gamma_{BD}=\delta^A_D$ $$\gamma^{AB},_B\gamma_{BD}+\gamma^{AB}\gamma_{BD},_B=0 $$ multiplying by $\gamma^{CD}$ $$\gamma^{AB},_B \gamma^{CD}\gamma_{BD}+\gamma^{CD}\gamma^{AB}\gamma_{BD},_B=0 $$ $$\gamma^{AB},_B\delta^C_B+\gamma^{CD}\gamma^{AB}\gamma_{BD},_B=0$$ $C=B$ $$\gamma^{AB},_B+\gamma^{BD}\gamma^{AB}\gamma_{BD},_B=0$$ which is different from the wished result. What I did wrong?
Your guess is correct. If $\gamma^{AB}\gamma_{BC}=\delta^A_C$, then $$\gamma^{AB}_{\quad ,D}\gamma_{BC} + \gamma^{AB}\gamma_{BC,D}=0\implies \gamma^{AB}_{\quad,D}\gamma_{BC}=-\gamma^{AB}\gamma_{BC,D}.$$Multiply by $\gamma^{CE}$ on both sides to get $$\gamma^{AE}_{\quad,D}=\gamma^{AB}_{\quad,D}\delta_B^E = -\gamma^{AB}\gamma_{BC,D}\gamma^{CE}.$$Rename $E\leftrightarrow B$ and $C\leftrightarrow D$ to get $$\gamma^{AB}_{\quad,C} = -\gamma^{AE}\gamma_{DE,C}\gamma^{BD}.$$