Question
Differentiate the function $\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}},-2<x<2$ w.r.t to x.
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Answer
Let y = $\frac{\cos ^{-1} \frac x2}{\sqrt{2 x+7}},-2<x<2$
Differentiating both sides with respect to x, we get
Using Quotient rule
$\frac{d y}{d x}=\frac{\sqrt{2 x+7} \frac{d}{d x}\left(\cos ^{-1} \frac{x}{2}\right)-\left(\cos ^{-1} \frac{x}{2}\right) \frac{d}{d x}(\sqrt{2 x+7})}{(\sqrt{2 x+7})^{2}}$
$\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{x}}$ = $\frac{\sqrt{2 x+7}\left[\frac{-1}{\sqrt{1-\left(\frac{x}{2}\right)^{2}}} \times \frac{d}{d x}\left(\frac{x}{2}\right)\right]-\left(\cos ^{-1} \frac{x}{2}\right) \times \frac{1}{2 \sqrt{2 x+7}} \times \frac{d}{d x}(2 x+7)}{2 x+7}$
$\frac{d y}{d x}=\frac{\sqrt{2 x+7} \times-\frac{1}{\sqrt{4-x^{2}}}-\left(\cos ^{-1} \frac{x}{2}\right) \times \frac{2}{2 \sqrt{2 x+7}}}{2 x+7}$
$\frac{d y}{d x}=-\frac{\sqrt{2 x+7}}{\sqrt{4-x^{2}} \times(2 x+7)}-\frac{\cos ^{-1} \frac{x}{2}}{(\sqrt{2 x+7})(2 x+7)}$
$\therefore$ $\frac{d y}{d x}=-\left[\frac{1}{\sqrt{4-x^{2}} \times \sqrt{2 x+7}}+\frac{\cos ^{-1} \frac{x}{2}}{(2 x+7)^{\frac{3}{2}}}\right]$
Differentiating both sides with respect to x, we get
Using Quotient rule
$\frac{d y}{d x}=\frac{\sqrt{2 x+7} \frac{d}{d x}\left(\cos ^{-1} \frac{x}{2}\right)-\left(\cos ^{-1} \frac{x}{2}\right) \frac{d}{d x}(\sqrt{2 x+7})}{(\sqrt{2 x+7})^{2}}$
$\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{x}}$ = $\frac{\sqrt{2 x+7}\left[\frac{-1}{\sqrt{1-\left(\frac{x}{2}\right)^{2}}} \times \frac{d}{d x}\left(\frac{x}{2}\right)\right]-\left(\cos ^{-1} \frac{x}{2}\right) \times \frac{1}{2 \sqrt{2 x+7}} \times \frac{d}{d x}(2 x+7)}{2 x+7}$
$\frac{d y}{d x}=\frac{\sqrt{2 x+7} \times-\frac{1}{\sqrt{4-x^{2}}}-\left(\cos ^{-1} \frac{x}{2}\right) \times \frac{2}{2 \sqrt{2 x+7}}}{2 x+7}$
$\frac{d y}{d x}=-\frac{\sqrt{2 x+7}}{\sqrt{4-x^{2}} \times(2 x+7)}-\frac{\cos ^{-1} \frac{x}{2}}{(\sqrt{2 x+7})(2 x+7)}$
$\therefore$ $\frac{d y}{d x}=-\left[\frac{1}{\sqrt{4-x^{2}} \times \sqrt{2 x+7}}+\frac{\cos ^{-1} \frac{x}{2}}{(2 x+7)^{\frac{3}{2}}}\right]$
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