Derivative of a Definite Integral

I have two equations as follows,

$ $ a = \int_0^1 dx \frac{c z_s^{d+1} x^d}{\sqrt{(1-(z_s/z_h)^{d+1} x^{d+1})(1-c^2 z_s^{2d} x^{2d})}} \tag{1}\label{1},$ $

\begin{align} S &= \frac{1}{4 z_s^{d-1}}\Bigg(-\frac{\sqrt{(1-c^2 z_s^{2d})(1-b^{d+1})}}{d-1} – \frac{1}{d-1} c^2 z_s^{2d} \int^1_0 dx x^d \sqrt{\frac{(1-(b x)^{d+1})}{(1-c^2(z_s x)^{2d})}}\ & -\frac{b^{d+1}(d+1)}{2(d-1)} \int^1_0 dx x \sqrt{\frac{(1-c^2(z_s x)^{2d})}{(1-(b x)^{d+1})}}\ & + b^{d+1}\int^1_0 dx \frac{x}{\sqrt{(1-(b x)^{d+1})(1-c^2(z_s x)^{2d})}}\Bigg) \tag{2}\label{2} \end{align}

where $ c=c(z_s)$ is a function of $ z_s$ , while $ a$ (I can fix a value for this), $ z_h$ , $ d$ (dimension) are constants, also $ b=z_s/z_h$ .

My goal is to obtain an expression for $ S$ independent of $ c$ , the conditions that can help with this requirement are,

$ $ \frac{dS}{dz_s} = 0 \tag{3}\label{3},$ $

and $ \eqref{1}$ . From $ \eqref{1}$ and $ \eqref{2}$ , we can take the derivative with respect to $ z_s$ and impose $ \eqref{3}$ on the derivative of $ \eqref{2}$ so that we can get expressions involving $ c$ and $ \frac{dc}{dz_s}$ in both the derivatives of $ \eqref{1}$ and $ \eqref{2}$ , then eliminate both $ c$ and $ \frac{dc}{dz_s}$ .

However, I tried to do the typical way as in the documentation but it does not produce the result I want.

d = 3; Integrate[(c[zs] zs^(d + 1) x^d)/((1 - x^(d + 1) (zs/zh)^(d + 1)) (1 - c[zs]^2 (zs x)^(2 d)))^(1/2), {x, 0, 1}] D[Integrate[(c[zs] zs^(d + 1) x^d)/((1 - x^(d + 1) (zs/zh)^(d + 1)) (1 - c[zs]^2 (zs x)^(2 d)))^(1/2), {x, 0, 1}] == 0, zs]