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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Differentiate w.r.t. x the function in Exercise:
$\frac{\cos^{-1}\frac{\text{x}}{2}}{\sqrt{2\text{x}+7}},-2<\text{x}<2$
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Answer

Let $\text{y}=\frac{\cos^{-1}\frac{\text{x}}{2}}{\sqrt{2\text{x}+7}}$By quotient rule, we obtain
$\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{2\text{x}+7}\frac{\text{d}}{\text{dx}}\Big(\cos^{-1}\frac{\text{x}}{2}\Big)-\Big(\cos^{-1}\frac{\text{x}}{2}\Big)\frac{\text{d}}{\text{dx}}(\sqrt{2\text{x}+7)}}{(\sqrt{2\text{x}+7})^2}$
$=\frac{\sqrt{2\text{x}+7}\Bigg[\frac{1}{\sqrt{1-\Big(\frac{\text{x}}{2}\Big)^2}}.\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{2}\Big)\Bigg]-\Big(\cos^{-1}\frac{\text{x}}{2}\Big)\frac{1}{2\sqrt{2\text{x}+7}}.\frac{\text{d}}{\text{dx}}(2\text{x}+7)}{2\text{x}+7}$
$=\frac{\sqrt{2\text{x}+7}\frac{-1}{\sqrt{4-\text{x}^2}}-\Big(\cos^{-1}\frac{\text{x}}{2}\Big)\frac{2}{2\sqrt{2\text{x}+7}}}{2\text{x}+7}$
$=\frac{-\sqrt{2\text{x}+7}}{\sqrt{4-\text{x}^2}\times(2\text{x}+7)}-\frac{\cos^{-1}\frac{\text{x}}{2}}{(\sqrt{2\text{x}+7})(2\text{x}+7)}$
$=-\Bigg[\frac{1}{\sqrt{4-\text{x}^2}\sqrt{2\text{x}+7}}+\frac{\cos^{-1}\frac{\text{x}}{2}}{(2\text{x}+7)^{\frac{3}{2}}}\Bigg]$

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