Prove that cosec'(x) = -cosec(x)*cot(x)

Start Function Definition: Step 1: Start by writing the function whose derivative we need to find. y = cosec(x) = 1/sin(x)Apply Chain Rule: Step 2: Use the chain rule to differentiate y with respect to x. y' = -(1/sin^2(x)) * cos(x)Simplify Using Identities: Step 3: Simplify the expression using trigonometric identities. y' = -cosec(x) * cot(x)

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Prove that $\csc'(x) = -\csc(x)\cot(x)$

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Algebra 1

Simplify exponential expressions using exponent rules

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Q. Prove that

\csc'(x) = -\csc(x)\cot(x)

Start Function Definition: Step $1$ : Start by writing the function whose derivative we need to find. $\newline$ $y = \csc(x) = \frac{1}{\sin(x)}$
Apply Chain Rule: Step $2$ : Use the chain rule to differentiate $y$ with respect to $x$ . $y' = -\left(\frac{1}{\sin^2(x)}\right) \cdot \cos(x)$
Simplify Using Identities: Step $3$ : Simplify the expression using trigonometric identities. $\newline$ $y' = -\csc(x) \cdot \cot(x)$