if the

Let $y=\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}$

By quotient rule, we obtain

$\frac{d y}{d x}=\frac{\sqrt{2 x+7} \frac{d}{d x}\left(\cos ^{-1} \frac{x}{2}\right)-\left(\cos ^{-1} \frac{x}{2}\right) \frac{d}{d x}(\sqrt{2 x+7})}{(\sqrt{2 x+7})^{2}}$

$=\frac{\sqrt{2 x+7}\left[\frac{-1}{\sqrt{1-\left(\frac{x}{2}\right)^{2}}} \cdot \frac{d}{d x}\left(\frac{x}{2}\right)\right]-\left(\cos ^{-1} \frac{x}{2}\right) \frac{1}{2 \sqrt{2 x+7}} \cdot \frac{d}{d x}(2 x+7)}{2 x+7}$

$=\frac{\sqrt{2 x+7} \frac{-1}{\sqrt{4-x^{2}}}-\left(\cos ^{-1} \frac{x}{2}\right) \frac{2}{2 \sqrt{2 x+7}}}{2 x+7}$

$=\frac{-\sqrt{2 x+7}}{\sqrt{4-x^{2}} \times(2 x+7)}-\frac{\cos ^{-1} \frac{x}{2}}{(\sqrt{2 x+7})(2 x+7)}$

$=-\left[\frac{1}{\sqrt{4-x^{2}} \sqrt{2 x+7}}+\frac{\cos ^{-1} \frac{x}{2}}{(2 x+7)^{\frac{3}{2}}}\right]$