Cosec(x) . Cot(2x)
Derivatives
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Answer:
∴ d y d x = d d x [ c o s e c x c o t x ] dydx=ddx[cosecxcotx] = ( d d x c o s e c x ) c o t x + c o s e c x d d x c o t x =(ddxcosecx)cotx+cosecxddxcotx = cot x ∙(− cot x cosec x) + cosec x (- cosec2x) = − cot2x cosec − cosec3x = −cosec x (cot2x + cosec2x)
Step-by-step explanation:
∴ d y d x = d d x [ c o s e c x c o t x ] dydx=ddx[cosecxcotx] = ( d d x c o s e c x ) c o t x + c o s e c x d d x c o t x =(ddxcosecx)cotx+cosecxddxcotx = cot x ∙(− cot x cosec x) + cosec x (- cosec2x) = − cot2x cosec − cosec3x = −cosec x (cot2x + cosec2x)
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