Question

# Differentiate the given function w.r.t. $$x$$:$$\displaystyle{\frac{\cos^{-1} \dfrac{x}{2}}{\sqrt{2x + 7}}}$$, $$-2 < x < 2$$

Solution

## Let  $$y =\displaystyle{\frac{\cos^{-1} \frac{x}{2}}{\sqrt{2x + 7}}}$$Thus using  quotient rule, we obtain $$\displaystyle{\frac{dy}{dx} = \frac{\sqrt{2x + 7}\dfrac{d}{dx}\left(\cos^{-1}\frac{x}{2}\right) - \left(\cos^{-1}\frac{x}{2}\right) \dfrac{d}{dx}(\sqrt{2x + 7})}{(\sqrt{2x + 7})^2}}$$ $$= \displaystyle{\dfrac{\sqrt{2x + 7} \left[\dfrac{-1}{\sqrt{1 - \left(\frac{x}{2}\right)^{2}}}.\dfrac{d}{dx}\left(\dfrac{x}{2}\right)\right] - \left(\cos^{-1}\frac{x}{2}\right).\dfrac{1}{2\sqrt{2x + 7}}\dfrac{d}{dx}(2x + 7)}{2x + 7}}$$$$= \displaystyle{\frac{\sqrt{2x +7} \dfrac{-1}{\sqrt{4 - x^2}} - \left(\cos^{-1}\frac{x}{2}\right)\dfrac{1}{2\sqrt{2x + 7}}\times 2}{2x + 7}}$$ $$\displaystyle = -\left[\frac{1}{\sqrt{4 - x^2} \sqrt{2x + 7}} + \frac{\cos^{-1} \frac{x}{2}}{(2x + 7)^{\frac{3}{2}}}\right]$$ MathematicsRS AgarwalStandard XII

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