Variation of induced metric in Nambu-Goto action I'm working with the Nambu-Goto action
$$
S=-\mu\int d^2\zeta\sqrt{\gamma}
$$
with $\gamma$ the determinant of the pull-back metric
$$
\gamma_{ab}=
\begin{pmatrix}
\dot{X}^2 & \dot{X}\cdot X' \\
\dot{X}\cdot X' & X'^2
\end{pmatrix}
=\frac{dX^{\mu}}{d\zeta^a}\frac{dX^{\nu}}{d\zeta^b}\eta_{\mu\nu}
$$
where dots and primes relate to differentiation with respect to the components of the parameterization of $\zeta=\zeta(\tau,\sigma).$ In his string theory notes, Tong suggests that there is a straightforward way to get the equations of motion of this system. Namely
$$
0=\delta S=-\mu\int d^2\zeta\delta\sqrt{-\gamma}
$$
makes Euler-Lagrange equations. We can exploit a well-known identity that
$$
\delta\sqrt{-\gamma}=\frac{1}{2}\sqrt{-\gamma}\gamma^{ab}\delta\gamma_{ab}.
$$
Of course this must vanish via the principle of stationary action. Tong (and many others) gives the resulting Euler-Lagrange equation from this variation as
$$
\partial_{\alpha}(\sqrt{-\gamma}\gamma^{ab}\partial_{\beta}X^{\mu})=0.
$$
Obviously the variation of the pullback must be $2\partial_{\beta}X^{\mu}$, but aside from the factor of two (index manipulation) I'm not sure why this is the right answer. This question has been asked before without a satisfactory answer.
 A: $$
\gamma_{ab} = \partial_a X^\mu \partial_b X_\mu \implies \gamma^{ab} \delta \gamma_{ab}  = 2 \gamma^{ab} \partial_a X^\mu \partial_b \delta X_\mu.
$$
Then,
\begin{align}
\delta S &= - \frac{\mu}{2} \int d^2 \zeta  \sqrt{-\gamma}2 \gamma^{ab} \partial_a X^\mu \partial_b \delta X_\mu \\
&=  \mu \int d^2 \zeta \partial_b [ \sqrt{-\gamma} \gamma^{ab} \partial_a X^\mu ] \delta X_\mu + \text{bdy term}
\end{align}
The equation of motion is then
$$
 \partial_b [ \sqrt{-\gamma} \gamma^{ab} \partial_a X^\mu ] = 0 \implies \nabla^2 X^\mu = 0.
$$

Variations
OP asked me to clarify how the variation of $\gamma$ was derived. Think of $\gamma_{ab}$ as a function of $X$,
$$
\gamma_{ab}(X) = \partial_a X^\mu \partial_b X_\mu
$$
Then, the variation is the simple statement
$$
\delta \gamma_{ab}(X) = \gamma_{ab} ( X + \delta X ) - \gamma_{ab} (X)
$$
where on the RHS, only the linear term in $\delta X$ is kept.
In our case, we have
\begin{align}
\delta \gamma_{ab}(X) &= \partial_a [ X^\mu + \delta X^\mu ] \partial_b [ X_\mu + \delta X^\mu ] - \partial_a X^\mu \partial_b X_\mu \\
&= \partial_a X^\mu \partial_b \delta X_\mu + \partial_a \delta X^\mu \partial_b X_\mu + O( (\delta X)^2 ) . 
\end{align}
It follows that
$$
\gamma^{ab} \delta \gamma_{ab}  = 2 \gamma^{ab} \partial_a X^\mu \partial_b \delta X_\mu
$$
