Ex 5.2, 7 - Differentiate 2 root cot(x2) - Chapter 5 NCERT - Finding derivative of a function by chain rule

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.2, 7 (Introduction) Differentiate the functions with respect to π‘₯ 2√(cot ( π‘₯2 )) Finding derivative of cot x Let 𝑦=cot⁑π‘₯ 𝑦= (π‘π‘œπ‘  π‘₯)/(𝑠𝑖𝑛 π‘₯) Let 𝑒 = cos⁑π‘₯ & 𝑣=sin π‘₯ 𝑦′ = (𝑒/𝑣)^β€² = (𝑒^β€² 𝑣 βˆ’ γ€– 𝑣〗^β€² 𝑒)/𝑣^2 Now, 𝑒 = cos⁑π‘₯ & 𝑣=sin π‘₯ 𝑒′ = βˆ’sin π‘₯ & 𝑣′=cos⁑π‘₯ Thus, 𝑦 = (𝑒^β€² 𝑣 βˆ’ γ€– 𝑣〗^β€² 𝑒)/𝑣^2 = (βˆ’sin π‘₯" " sin π‘₯" " βˆ’ cos⁑π‘₯ " " cos⁑π‘₯)/(sin^2 π‘₯) = (βˆ’(sin^2 π‘₯" +" cos^2⁑π‘₯))/(sin^2 π‘₯) = (βˆ’1)/(sin^2 π‘₯) = βˆ’cosec^2⁑π‘₯ Thus, 𝑦′ = (𝑒^β€² 𝑣 βˆ’ γ€– 𝑣〗^β€² 𝑒)/𝑣^2 = (βˆ’sin π‘₯" " sin π‘₯" " βˆ’ cos⁑π‘₯ " " cos⁑π‘₯)/(sin^2 π‘₯) = (βˆ’(sin^2 π‘₯" +" cos^2⁑π‘₯))/(sin^2 π‘₯) = (βˆ’1)/(sin^2 π‘₯) = βˆ’cosec^2⁑π‘₯ Ex 5.2, 7 Differentiate the functions with respect to π‘₯ 2√(cot ( π‘₯2 ))

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