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Continuity And Differentiability

Question
CBSEENMA12035285

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.

Solution
Here 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
space space space space space space space space space space space space equals tan to the power of negative 1 end exponent open parentheses fraction numerator 5 xx minus straight x over denominator 1 plus 5 straight x. straight x end fraction close parentheses plus tan to the power of negative 1 end exponent open parentheses fraction numerator straight x plus begin display style 2 over 3 end style over denominator 1 minus straight x. begin display style 2 over 3 end style end fraction close parentheses
space space space space space space space space space space space space equals tan to the power of negative 1 end exponent 5 straight x minus tan to the power of negative 1 end exponent straight x plus tan to the power of negative 1 end exponent 2 over 3 equals tan to the power of negative 1 end exponent 5 straight x plus tan to the power of negative 1 end exponent 2 over 3
therefore space dy over dx equals fraction numerator 5 over denominator 1 plus left parenthesis 5 straight x right parenthesis squared end fraction plus 0 equals fraction numerator 5 over denominator 1 plus 25 straight x squared end fraction

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