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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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Here 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 equals tan to the power of negative 1 end exponent 1 minus tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent 2 equals tan to the power of negative 1 end exponent 1 plus tan to the power of negative 1 end exponent 2 therefore space dy over dx equals 0 plus 0 space rightwards double arrow 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.

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