Question:
Differentiate w.r.t. \(x\) the function: \(\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}},-2<x<2\)
Differentiate w.r.t. \(x\) the function: \(\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}},-2<x<2\)
Updated On: Oct 19, 2023
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Solution and Explanation
The correct answer is \(=-\bigg[\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}}+\frac{cos^{-1}\frac{x}{2}}{(2x+7)^{\frac{3}{2}}}\bigg]\)
Let \(y=\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}\)
By quotient rule,we obtain
\(\frac{dy}{dx}=\frac{\sqrt{2x+7}\frac{d}{dx}(cos^{-1}\frac{x}{2})-(cos^{-1}\frac{x}{2})\frac{d}{dx}(\sqrt{2x+7})}{\sqrt{(2x+7)^2}}\)
\(=\frac{\sqrt{2x+7}\bigg[\frac{-1}{\sqrt{1-(\frac{x}{2})^2}}.\frac{d}{dx}(\frac{x}{2})\bigg]-(cos^{-1}\frac{x}{2})\frac{1}{2\sqrt{2x+7}}.\frac{d}{dx}(2x+7)}{2x+7}\)
\(=\frac{\sqrt{2x+7}\frac{-1}{\sqrt{4-x^2}}-(cos^{-1}\frac{x}{2})\frac{2}{2\sqrt{2x+7}}}{2x+7}\)
\(=\frac{-\sqrt{2x+7}}{\sqrt{4-x^2}(2x+7)}-\frac{cos^{-1}\frac{x}{2}}{(\sqrt{2x+7})(2x+7)}\)
\(=-\bigg[\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}}+\frac{cos^{-1}\frac{x}{2}}{(2x+7)^{\frac{3}{2}}}\bigg]\)
Let \(y=\frac{cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}\)
By quotient rule,we obtain
\(\frac{dy}{dx}=\frac{\sqrt{2x+7}\frac{d}{dx}(cos^{-1}\frac{x}{2})-(cos^{-1}\frac{x}{2})\frac{d}{dx}(\sqrt{2x+7})}{\sqrt{(2x+7)^2}}\)
\(=\frac{\sqrt{2x+7}\bigg[\frac{-1}{\sqrt{1-(\frac{x}{2})^2}}.\frac{d}{dx}(\frac{x}{2})\bigg]-(cos^{-1}\frac{x}{2})\frac{1}{2\sqrt{2x+7}}.\frac{d}{dx}(2x+7)}{2x+7}\)
\(=\frac{\sqrt{2x+7}\frac{-1}{\sqrt{4-x^2}}-(cos^{-1}\frac{x}{2})\frac{2}{2\sqrt{2x+7}}}{2x+7}\)
\(=\frac{-\sqrt{2x+7}}{\sqrt{4-x^2}(2x+7)}-\frac{cos^{-1}\frac{x}{2}}{(\sqrt{2x+7})(2x+7)}\)
\(=-\bigg[\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}}+\frac{cos^{-1}\frac{x}{2}}{(2x+7)^{\frac{3}{2}}}\bigg]\)
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