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Continuity And Differentiability

Question
CBSEENMA12035394
Wired Faculty App

If space straight y equals straight a space straight e to the power of mx plus straight b space straight e to the power of negative mx end exponent comma space prove space that space fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight m squared space straight y equals 0

Solution
space space space space straight y equals straight a space straight e to the power of mx plus straight b space straight e to the power of negative mx end exponent 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 space space space space space space space space space space... left parenthesis 1 right parenthesis therefore space space dy over dx equals straight a space straight e to the power of mx space. straight m plus straight b space straight e to the power of negative mx end exponent left parenthesis negative straight m right parenthesis equals straight m left parenthesis straight a space straight e to the power of mx minus straight b space straight e to the power of negative mx end exponent right parenthesis therefore space fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight m left square bracket straight a space straight e to the power of mx. straight m minus straight b space straight e to the power of negative mx end exponent left parenthesis negative straight m right parenthesis right square bracket equals straight m squared space left square bracket straight a space straight e to the power of mx plus straight b space straight e to the power of negative mx end exponent right square bracket therefore space fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight m squared straight y 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 space space space space space space space space space space space space space space space left square bracket because space of space 1 right square bracket therefore space fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight m squared straight y equals 0

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