# 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]