The following is from Huybrechts' Complex Geometry - An Introduction, chapter 6.
Let $X$ be a complex manifold with complex structure $I_0$, and let $I_t, t \in \mathbb C$ be a 1-parameter family of complex structures. For $t$ small this defines a map $$ \phi(t): T^{0,1} \to T^{1,0},$$ such that $T^{0,1}_t = \{u + \phi(t)(u) \,|\, u \in T^{0,1}\}$, where $T_{X, \mathbb C} = T^{1,0}_t \oplus T^{0,1}_t$ is the decomposition associated to the complex structure $I_t$. Let's develop $\phi(t)$ as a power series $$ \phi(t) = \phi_1 t + \phi_2 t^2 + \dotsc, $$ for $\phi_i \in A^{0,1}(\mathcal T_X)$. The integrability of $I_t$ is equivalent to the Maurer-Cartan equation $$ \bar \partial \phi(t) + \frac{1}{2} [\phi(t), \phi(t)] = 0,$$ and if we split this by degrees of $t$ we get a system of equations \begin{align*} \bar \partial \phi_1 & = 0 \\ \bar \partial \phi_2 + \frac{1}{2} [\phi_1, \phi_1] & = 0 \\ \vdots \end{align*} In particular we see that $\phi_1$ is $\bar \partial$-closed, and we call the Dolbeault cohomology class $[\phi_1] \in H^1(X, \mathcal T_X)$ the Kodaira-Spencer class of $I_t$.
Huybrechts then claims
any class in $H^1(X,\mathcal T_X)$ occurs as a Kodaira-Spencer class.
Why is that true? By the Maurer-Cartan equations, we need at least that $[\phi_1, \phi_1] \in A^{0,2}(\mathcal T_X)$ is $\bar \partial$-exact, but why is that the case? And even if this is true, we still need to show that the higher order equations are solvable...