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The Derivative of Sec − 1 ( 1 2 X 2 + 1 ) W . R . T . √ 1 + 3 X at X = − 1 / 3 - Mathematics

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Question

The derivative of \[\sec^{- 1} \left( \frac{1}{2 x^2 + 1} \right) \text { w . r . t }. \sqrt{1 + 3 x} \text { at } x = - 1/3\]

Options

  • does not exist

  • 0

  • 1/2

  • 1/3

MCQ

Solution

does not exist

\[\text { We know that } \sec^{- 1} \alpha\text { is not defined for }\alpha \in \left( - 1, 1 \right)\]
\[\text { Here for } x = \frac{- 1}{3}, \frac{1}{2 x^2 + 1} = \frac{9}{11} \in \left( - 1, 1 \right)\]
\[ \therefore \sec^{- 1} \left( \frac{1}{2 x^2 + 1} \right)\text { is not defined at } x = \frac{- 1}{3}\]
\[ \therefore \text { Derivative of } \sec^{- 1} \left( \frac{1}{2 x^2 + 1} \right) \text { does not exist at } x = \frac{- 1}{3}\]

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Chapter 11: Differentiation - Exercise 11.10 [Page 120]

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RD Sharma Mathematics [English] Class 12
Chapter 11 Differentiation
Exercise 11.10 | Q 11 | Page 120

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