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Continuity and Differentiability

Short Answer Type

481.

Prove space that space dy over dx space is space independent space of space straight x comma space when straight y equals cot to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus sin space straight x end root plus square root of 1 minus sin space straight x end root over denominator square root of 1 plus sin space straight x end root minus square root of 1 minus sin space straight x end root end fraction close square brackets comma space open parentheses 0 less than straight x less than straight pi over 2 close parentheses.

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482.

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

91 Views

483.

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

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484. If y = tan^-1

open parentheses fraction numerator straight x to the power of begin display style 1 third end style end exponent plus straight a to the power of begin display style 1 third end style end exponent over denominator 1 minus straight x to the power of begin display style 1 third end style end exponent space straight a to the power of begin display style 1 third end style end exponent end fraction close parentheses

when 0 < a x < 1, prove that

dy over dx

is independent of a.

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485.

straight y equals sin to the power of negative 1 end exponent open curly brackets straight x square root of 1 minus straight x end root minus square root of straight x square root of 1 minus straight x squared end root close curly brackets space semicolon space find space dy over dx comma

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486.

If space straight y equals sin to the power of negative 1 end exponent left parenthesis straight x squared square root of 1 minus straight x squared end root plus straight x square root of 1 minus straight x to the power of 4 end root right parenthesis comma space show space that space dy over dx minus fraction numerator 2 straight x over denominator square root of 1 minus straight x to the power of 4 end root end fraction equals fraction numerator 1 over denominator square root of 1 minus straight x to the power of 4 end root end fraction

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487.

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

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488.

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

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489.

Differentiate space tan to the power of negative 1 end exponent open parentheses fraction numerator 3 straight x minus straight x cubed over denominator 1 minus 3 straight x squared end fraction close parentheses space straight w. straight r. straight t. space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction close parentheses

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490. $Differentiate space colon space cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses space straight w. straight r. straight t. space tan to the power of negative 1 end exponent open parentheses fraction numerator 3 straight x minus straight x cubed over denominator 1 minus 3 straight x squared end fraction close parentheses$

Let space space space space space straight y equals cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses Put space space space space space straight x equals tan space straight theta therefore space space space space space space space straight y equals cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus tan squared space straight theta over denominator 1 plus tan squared space straight theta end fraction close parentheses space space space space space space space space space space space space space equals cos to the power of negative 1 end exponent left parenthesis cos space 2 straight theta right parenthesis equals 2 straight theta therefore space space space space space space space straight y equals 2 tan to the power of negative 1 end exponent space straight x therefore space space dy over dx equals fraction numerator 2 over denominator 1 plus straight x squared end fraction Also space space space space straight u equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 3 straight x minus straight x cubed over denominator 1 minus 3 straight x squared end fraction close parentheses Put space space space space space space space straight x equals space tan space straight theta therefore space space space space space space space space space space space straight u equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 3 tan space straight theta minus tan cubed space straight theta over denominator 1 minus 3 tan squared space straight theta end fraction close parentheses space space space space space space space space space space space space space space space space space equals tan to the power of negative 1 end exponent left parenthesis tan space 3 straight theta right parenthesis equals 3 straight theta therefore space space space space space space space space space space space straight u equals 3 space tan to the power of negative 1 end exponent space straight x therefore space space space space space du over dx equals fraction numerator 3 over denominator 1 plus straight x squared end fraction Now space dy over dx equals fraction numerator begin display style dy over dx end style over denominator begin display style du over dx end style end fraction equals fraction numerator 2 over denominator 1 plus straight x squared end fraction cross times fraction numerator 1 plus straight x squared over denominator 3 end fraction therefore space dy over dx equals 2 over 3

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