Differentiate with respect to the functions in exercise contexto.140 3 years ago Solution: Now let’s consider Derivating the above function:
![\begin{array}{l} \frac{d y}{d x}=\frac{-1}{\left.\sqrt{1-\left(\frac{x}{2}\right)^{2}}\right) \frac{d}{d x} \frac{x}{2}-\left(\cos ^{-1} \frac{x}{2}\right) \frac{1}{2}(2 x+7)^{\frac{-1}{2}} \frac{d}{d x}(2 x+7)}{(\sqrt{2 x+7})^{2}} \\ \begin{array}{l} \frac{d y}{d x}=\frac{-\sqrt{2 x+7} \cdot \frac{2}{\sqrt{4-x^{2}}} \cdot \frac{1}{2}-\frac{1}{2} \cdot \frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}}{(2 x+7)} \\\end{array} \\ \frac{\left.-\frac{\sqrt{2 x+7}}{\sqrt{4-x^{2}}}+\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}\right]}{(2 x+7)} \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-631d54d9713e8b835e0e790e48877628_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} \frac{d y}{d x}=-\left[\frac{2 x+7+\sqrt{4-x^{2}} \cos ^{-1} \frac{x}{2}}{\sqrt{4-x^{2}} \sqrt{2 x+7}(2 x+7)}\right] \\ =\left[\frac{2 x+7+\sqrt{4-x^{2}} \cos ^{-1} \frac{x}{2}}{\sqrt{4-x^{2}}(2 x+7)^{\frac{3}{2}}}\right] \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ae874fe7aa3edc88a735347ffec12770_l3.png "Rendered by QuickLaTeX.com")
