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Continuity and Differentiability

Short Answer Type

551. If y=e^{a x}sin bx, prove that

fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 straight a dy over dx plus left parenthesis straight a squared plus straight b squared right parenthesis straight y equals 0

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552. $If space straight y equals straight x space log open parentheses fraction numerator straight x over denominator straight a plus straight b space straight x end fraction close parentheses comma space prove space that space straight x cubed fraction numerator straight d squared straight y over denominator dx squared end fraction equals open parentheses straight x dy over dx minus straight y squared close parentheses.$

Here space space space straight y equals straight x space log open parentheses fraction numerator straight x over denominator straight a plus straight b space straight x end fraction close parentheses therefore space space space space space straight y over straight x equals log space straight x minus log left parenthesis straight a plus straight b space straight x right parenthesis Differentaiting space straight w. straight r. tx comma space we space get comma space space space space space fraction numerator straight x begin display style dy over dx end style minus straight y.1 over denominator straight x squared end fraction equals 1 over straight x minus fraction numerator straight b over denominator straight a plus straight b space straight x end fraction space space space space space space space straight x dy over dx minus straight y equals straight x minus straight b space straight x squared straight a plus straight b space straight x therefore space space space straight x dy over dx minus straight y equals fraction numerator straight a space straight x plus straight b space straight x squared minus straight b space straight x squared over denominator straight a plus straight b space straight x end fraction therefore space space space straight x dy over dx minus straight y equals fraction numerator straight a space straight x over denominator straight a plus straight b space straight x end fraction 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 space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis Again space differentiating space straight w. straight r. straight t. straight x comma space we space get comma space space space space space space straight x fraction numerator straight d squared straight y over denominator dx squared end fraction plus dy over dx.1 minus dy over dx equals fraction numerator left parenthesis straight a plus straight b space straight x right parenthesis. straight a space minus straight a space straight x. space straight b over denominator left parenthesis straight a plus straight b space straight x right parenthesis squared end fraction therefore space space space space straight x fraction numerator straight d squared straight y over denominator dx squared end fraction equals fraction numerator straight a squared over denominator left parenthesis straight a plus straight b space straight x right parenthesis squared end fraction rightwards double arrow space space straight x cubed fraction numerator straight d squared straight y over denominator dx squared end fraction equals fraction numerator straight a squared straight x squared over denominator left parenthesis straight a plus straight b space straight x right parenthesis squared end fraction rightwards double arrow space space straight x cubed fraction numerator straight d squared straight y over denominator dx squared end fraction equals open parentheses straight x dy over dx minus straight y close parentheses squared 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 space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket

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553.

If space straight x equals tan open parentheses 1 over straight a log space straight y close parentheses comma space show space that space left parenthesis 1 plus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction plus left parenthesis 2 space straight x minus straight a right parenthesis dy over dx equals 0.

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554. If y=x^x, prove that

fraction numerator straight d squared straight y over denominator dx squared end fraction minus 1 over straight y open parentheses dy over dx close parentheses squared minus straight y over straight x equals 0.

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555.

If space space space space straight y equals left parenthesis straight a plus straight b space straight t right parenthesis space straight e to the power of straight n space straight t end exponent comma space prove space that space fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 straight n dy over dx plus straight n squared straight y equals 0.

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556.

If space straight y equals fraction numerator straight a space straight x plus straight b over denominator straight c space straight x plus straight d end fraction comma space prove space that space 2 straight y subscript 1 space straight y subscript 3 equals 3 space straight y subscript 2 squared.

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557.

If space straight y equals log space left parenthesis 1 plus cos space straight x right parenthesis comma space prove space that space straight y subscript 1 space straight y subscript 2 plus straight y subscript 3 equals 0.

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558.

If space 2 space straight y equals straight x open parentheses 1 plus dy over dx close parentheses comma space show space that space fraction numerator straight d squared straight y over denominator dx squared end fraction space is space constant space.

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559. If y = (tan^-1 x)², then prove that (1 + x²)² y₂ + 2x(1 + x²)y_{1 = 2.}

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560.

If space straight y equals sin to the power of negative 1 end exponent straight x comma space then space show space that space open parentheses 1 minus straight x squared close parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight x dy over dx equals 0.

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