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Prove that csc⁡′(x)=−csc⁡(x)cot⁡(x)\csc'(x) = -\csc(x)\cot(x)

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Full solution

Q. Prove that csc⁡′(x)=−csc⁡(x)cot⁡(x)\csc'(x) = -\csc(x)\cot(x)
  1. Start Function Definition: Step 11: Start by writing the function whose derivative we need to find.\newliney=csc⁡(x)=1sin⁡(x)y = \csc(x) = \frac{1}{\sin(x)}
  2. Apply Chain Rule: Step 22: Use the chain rule to differentiate yy with respect to xx.y′=−(1sin⁡2(x))⋅cos⁡(x)y' = -\left(\frac{1}{\sin^2(x)}\right) \cdot \cos(x)
  3. Simplify Using Identities: Step 33: Simplify the expression using trigonometric identities.\newliney′=−csc⁡(x)⋅cot⁡(x)y' = -\csc(x) \cdot \cot(x)

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