The derivative of cosec x is
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
\(\csc x = \frac{1}{\sin x}\)
Using the quotient rule,
\(\frac{\mathrm d y}{\mathrm d x} = \frac{vdu - udv}{v^{2}}\)
= \(\frac{\sin x (0) - 1 (\cos x)}{(\sin x)^{2}}\)
= \(\frac{- \cos x}{\sin^{2} x}\)
= \((\frac{- \cos x}{\sin x}) (\frac{1}{\sin x})\)
= \(- \cot x \csc x\)
\(\csc x = \frac{1}{\sin x}\)
Using the quotient rule,
\(\frac{\mathrm d y}{\mathrm d x} = \frac{vdu - udv}{v^{2}}\)
= \(\frac{\sin x (0) - 1 (\cos x)}{(\sin x)^{2}}\)
= \(\frac{- \cos x}{\sin^{2} x}\)
= \((\frac{- \cos x}{\sin x}) (\frac{1}{\sin x})\)
= \(- \cot x \csc x\)