Application of flat base change

Application of flat base change

Let $f:X\rightarrow S$ be a flat morphism of schemes of finite type over
$\mathbb{C}$ such that both $S$ and the fibers of $f$ have rational
singularities.
Let $\varphi:S'\rightarrow S$ be a resolution of singularities and let
$X'=X\times_S S'$ be the fiber product, with projections $\varphi'$ and
$f'$ to $X$ and $S'$ respectively. Also, let $\psi:Y\rightarrow X'$ be a
resolution of singularities.
By assumption, we have that
$R^{\bullet}\psi\mathcal{O}_y=\mathcal{O}_{X'}$ and
$R^{\bullet}\varphi\mathcal{O}_{S'}=\mathcal{O}_S$. I am trying to show
that then $X$ also has rational singularities, namely
$R^{\bullet}(\varphi'\circ \psi)\mathcal{O}_Y=\mathcal{O}_X$.
By Grothendieck spectral sequence we have that $R^i(\varphi'\circ
\psi)\mathcal{O}_Y = R^i\varphi'\mathcal{O}_{X'}$ so we have to show that
$R^{\bullet}\varphi'\mathcal{O}_{X'}=\mathcal{O}_X$.
I am not sure about how this follows: by flat base change the natural map
$$f^{\ast}R^i\varphi_{\ast}\mathcal{O}_{S'}\rightarrow
R^i\varphi'f'^{\ast}\mathcal{O}_{S'}$$ is an isomorphism.
Now for instance for $i=0$ flat base change yields
$\varphi_{\ast}'f'^{\ast}\mathcal{O}_{S'}=f^{\ast}\mathcal{O}_S$. Taking
pushforward by $f$ on both sides yields
$$f_{\ast}\varphi'_{\ast}f'^{\ast}\mathcal{O}_{S'}=f_{\ast}f^{\ast}\mathcal{O}_S=\mathcal{O}_S\otimes
f_{\ast}\mathcal{O}_X=f_{\ast}\mathcal{O}_X$$
Commutativity on the LHS yields
$$f_{\ast}\varphi'_{\ast}f'^{\ast}\mathcal{O}_{S'}=\varphi_{\ast}f'_{\ast}f'^{\ast}\mathcal{O}_{S'}=\varphi_{\ast}(\mathcal{O}_{S'}\otimes
f'_{\ast}\mathcal{O}_{S'})=\varphi_{\ast}f'_{\ast}\mathcal{O}_{X'}$$
So $f_{\ast}\varphi'_{\ast}\mathcal{O}_{X'}=f_{\ast}\mathcal{O}_X$.
Does it follow that $\varphi'_{\ast}\mathcal{O}_{X'}=\mathcal{O}_X$? Am I
making this way too complicated?