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Continuity And Differentiability

Question
CBSEENMA12034846

Differentiate space square root of fraction numerator left parenthesis straight x minus 3 right parenthesis left parenthesis straight x squared plus 4 right parenthesis over denominator 3 straight x squared plus 4 straight x plus 5 end fraction end root straight w. straight r. straight t. straight x

Solution
Let space space straight y equals open square brackets fraction numerator left parenthesis straight x minus 3 right parenthesis left parenthesis straight x squared plus 4 right parenthesis over denominator 3 straight x squared plus 4 straight x plus 5 end fraction close square brackets to the power of 1 half end exponent
therefore space log space straight y equals log open square brackets fraction numerator left parenthesis straight x minus 3 right parenthesis left parenthesis straight x squared plus 4 right parenthesis over denominator 3 straight x squared plus 4 straight x plus 5 end fraction close square brackets to the power of 1 half end exponent
therefore space log space straight y equals 1 half log open square brackets fraction numerator left parenthesis straight x minus 3 right parenthesis left parenthesis straight x squared plus 4 right parenthesis over denominator 3 straight x squared plus 4 straight x plus 5 end fraction close square brackets
therefore space log space straight y equals 1 half log left square bracket left parenthesis straight x minus 3 right parenthesis left parenthesis straight x squared plus 4 right parenthesis right square bracket minus 1 half log space left parenthesis 3 straight x squared plus 4 straight x plus 5 right parenthesis
therefore space log space straight y equals 1 half log left parenthesis straight x minus 3 right parenthesis plus 1 half log left parenthesis straight x squared plus 4 right parenthesis minus 1 half left parenthesis 3 straight x squared plus 4 straight x plus 5 right parenthesis
Differentiating space both space sides space straight w. straight r. straight t. straight x comma space we space get comma
space 1 over straight y dy over dx equals fraction numerator 1 over denominator 2 left parenthesis straight x minus 3 right parenthesis end fraction plus fraction numerator 1 over denominator 2 left parenthesis straight x squared plus 4 right parenthesis end fraction.2 straight x minus fraction numerator 1 over denominator 2 left parenthesis 3 straight x squared plus 4 straight x plus 5 right parenthesis end fraction. left parenthesis 6 straight x plus 4 right parenthesis
therefore space space space dy over dx equals straight y over 2 open square brackets fraction numerator 1 over denominator straight x minus 3 end fraction plus fraction numerator 2 straight x over denominator straight x squared plus 4 end fraction minus fraction numerator 6 straight x plus 4 over denominator 3 straight x squared plus 4 straight x plus 5 end fraction close square brackets
therefore space space space dy over dx equals 1 half square root of fraction numerator left parenthesis straight x minus 3 right parenthesis left parenthesis straight x squared plus 4 right parenthesis over denominator 3 straight x squared plus 4 straight x plus 5 end fraction end root open square brackets fraction numerator 1 over denominator straight x minus 3 end fraction plus fraction numerator 2 straight x over denominator straight x squared plus 4 end fraction minus fraction numerator 6 straight x plus 4 over denominator 3 straight x squared plus 4 straight x plus 5 end fraction close square brackets

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