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Continuity and Differentiability

Short Answer Type

501.

Prove space that space the space derivative space of space sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 2 straight x squared minus 1 end fraction close parentheses comma space straight x greater than 0 space straight w. straight r. straight t. space square root of 1 minus straight x squared end root space is space equal space to space the space derivative space straight l subscript straight n space left parenthesis straight x squared right parenthesis space with space respect space to space straight x comma space left square bracket straight l subscript straight n left parenthesis straight x squared right parenthesis equals space log to the power of straight e. left parenthesis straight a squared right parenthesis right square bracket.

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502.

Differentiate space tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator 1 plus square root of 1 minus straight x squared end root end fraction close parentheses space straight w. straight r. straight t. space sin open parentheses 2 space cot to the power of negative 1 end exponent space square root of fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction end root close parentheses

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503.

If space straight y equals tan to the power of negative 1 end exponent fraction numerator 4 straight x over denominator 1 plus 5 straight x squared end fraction plus tan to the power of negative 1 end exponent fraction numerator 2 plus 3 straight x over denominator 3 minus 2 straight x end fraction comma space prove space that space dy over dx equals fraction numerator 5 over denominator 1 plus 25 straight x squared end fraction.

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504.

Given space that space straight y equals tan to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 2 over denominator 1 minus 2 straight x end fraction close parentheses comma space minus 1 less than straight x less than 1 half. space After space using space the space property space of space inverse space trigonometric space function comma space show space that space dy over dx equals 0

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505.

If space straight theta equals cos to the power of negative 1 end exponent open parentheses straight r over straight k close parentheses minus fraction numerator square root of straight k squared minus straight r squared end root over denominator straight r end fraction comma space find space dθ over dr

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506. Differentiate the following w.r.t.x:

straight x space to the power of sin to the power of negative 1 end exponent straight x end exponent

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507.

Differentiate space the space following space straight w. straight r. straight t. straight x colon space left parenthesis sin space straight x right parenthesis to the power of cos to the power of negative 1 end exponent straight x end exponent

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508. Differentiate the following w.r.t.x:

straight x to the power of cos to the power of negative 1 end exponent straight x end exponent

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509.

If space straight y equals tan to the power of negative 1 end exponent straight x over straight y comma space then space evaluate space dy over dx.

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510. $If space left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis plus straight y to the power of cot space straight x end exponent equals 1 comma space find space dy over dx.$

Here space left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis to the power of straight y plus straight y to the power of cot space straight x end exponent equals 1 Put space left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis equals straight u comma space straight y to the power of cot space straight x end exponent equals straight v therefore space straight u plus straight v equals 1 space space space space space space space space space space space space space space space space rightwards double arrow space du over dx plus dv over dx equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis Now space straight u equals left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis to the power of straight y space rightwards double arrow space log space straight u equals log left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis to the power of straight y space rightwards double arrow space log space straight v equals straight y. log left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis therefore space 1 over straight u du over dx equals straight y. fraction numerator 1 over denominator tan to the power of negative 1 end exponent straight x end fraction. fraction numerator 1 over denominator 1 plus straight x squared end fraction plus dy over dx. log space tan to the power of negative 1 end exponent straight x therefore space du over dx equals left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis to the power of straight y open square brackets fraction numerator straight y over denominator left parenthesis 1 plus straight x squared right parenthesis tan to the power of negative 1 end exponent straight x end fraction plus dy over dx. log space tan to the power of negative 1 end exponent straight x close square brackets space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis Also space straight v equals straight y to the power of cot space straight x end exponent space rightwards double arrow space log space straight v equals log space straight y to the power of cot space straight x end exponent space rightwards double arrow space log space straight v equals cot space straight x. log space straight y therefore space 1 over straight v dv over dx equals cot space straight x.1 over straight y dy over dx minus cosec squared straight x. space log space straight y therefore space dv over dx equals straight y to the power of cot space straight x end exponent open square brackets fraction numerator cot space straight x over denominator straight y end fraction dy over dx minus cosec squared straight x. log space straight y close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis

From (1), (2) and (3), we get,

left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis to the power of straight y open square brackets fraction numerator straight y over denominator left parenthesis 1 plus straight x squared right parenthesis tan to the power of negative 1 end exponent straight x end fraction plus dy over dx. log space tan to the power of negative 1 end exponent straight x close square brackets plus straight y to the power of cot space straight x end exponent open square brackets fraction numerator cot space straight x over denominator straight y end fraction dy over dx minus cosec squared straight x. log space straight y close square brackets equals 0 open square brackets left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis to the power of straight y. space log space tan to the power of negative 1 end exponent straight x plus straight y to the power of cot space straight x end exponent. fraction numerator cot space straight x over denominator straight y end fraction close square brackets dy over dx equals straight y to the power of cot space straight x end exponent cosec squared straight x space log space straight y minus fraction numerator straight y left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis to the power of straight y over denominator left parenthesis 1 plus straight x squared right parenthesis tan to the power of negative 1 end exponent straight x end fraction space space space space space space space space therefore space dy over dx equals fraction numerator straight y to the power of cot space straight x end exponent cosec squared straight x space log space straight y minus begin display style fraction numerator straight y left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis to the power of straight y over denominator left parenthesis 1 plus straight x squared right parenthesis tan to the power of negative 1 end exponent straight x end fraction end style over denominator left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis to the power of straight y. space log space tan to the power of negative 1 end exponent straight x plus straight y to the power of cot space straight x end exponent. begin display style fraction numerator cot space straight x over denominator straight y end fraction end style end fraction

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