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Continuity And Differentiability

Question
CBSEENMA12034705

Differentiate the following w.r.t.x:fraction numerator 3 over denominator left parenthesis 2 straight x squared plus 5 right parenthesis squared end fraction

Solution
Let space straight y equals fraction numerator 3 over denominator left parenthesis 2 straight x squared plus 5 right parenthesis squared end fraction equals 3 left parenthesis 2 straight x squared plus 5 right parenthesis to the power of negative 2 end exponent
therefore space dy over dx equals 3 left parenthesis negative 2 right parenthesis left parenthesis 2 straight x squared plus 5 right parenthesis to the power of negative 3 end exponent. straight d over dx left parenthesis 2 straight x squared plus 5 right parenthesis equals 3 left parenthesis negative 2 right parenthesis left parenthesis 2 straight x squared plus 5 right parenthesis to the power of negative 3 end exponent.4 straight x
space space space space space space space space space space space space space equals negative fraction numerator 24 straight x over denominator left parenthesis 2 straight x squared plus 5 right parenthesis cubed end fraction

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