class 12 maths continuity and differentiability

$2\sqrt {\cot ({x^2})}$

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📘 Continuity and Differentiability NCERT Ex.5.2 ,Q.7,Page 166 SA

$2\sqrt {\cot ({x^2})}$

Official Solution

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Let $y = 2\sqrt {\cot ({x^2})}$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = 2\cfrac{d}{{dx}}\sqrt {\cot ({x^2})} = 2 \cdot \cfrac{1}{2}{\{ \cot ({x^2})\} ^{\cfrac{{ - 1}}{2}}} \cdot \cfrac{d}{{dx}}\cot ({x^2})$

$= \cfrac{1}{{\sqrt {\cot ({x^2})} }}\{ - \cos e{c^2}({x^2})\} \cfrac{d}{{dx}}({x^2})$

$= \cfrac{1}{{\sqrt {\cot ({x^2})} }}\{ - \cos e{c^2}({x^2})\} (2x)$
$= \cfrac{{ - 2x\cos e{c^2}({x^2})}}{{\sqrt {\cot ({x^2})} }} = \cfrac{{ - 2\sqrt 2 x}}{{\sin {x^2}\sqrt {\sin 2{x^2}} }}$

therefore $\frac{dy}{dx}=\cfrac{{ - 2\sqrt 2 x}}{{\sin {x^2}\sqrt {\sin 2{x^2}} }}$

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