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Continuity And Differentiability

Question
CBSEENMA12035226
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If space straight y equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 5 straight a over denominator straight a squared minus 6 straight x squared end fraction close parentheses comma space prove space that space dy over dx equals fraction numerator 3 straight a over denominator straight a squared plus 9 straight x squared end fraction plus fraction numerator 2 straight a over denominator straight a squared plus 4 straight x squared end fraction

Solution
because space space space space space space space space straight y equals space tan to the power of negative 1 end exponent fraction numerator 5 straight a over denominator straight a squared minus 6 straight x squared end fraction therefore space space space space space space space space straight y equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 5 begin display style straight x over straight a end style over denominator 1 minus 6 open parentheses begin display style straight x over straight a end style close parentheses squared end fraction close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator 3 straight x over denominator straight a end fraction end style plus begin display style fraction numerator 2 straight x over denominator straight a end fraction end style over denominator 1 minus open parentheses begin display style fraction numerator 3 straight x over denominator straight a end fraction end style close parentheses open parentheses begin display style fraction numerator 2 straight x over denominator straight a end fraction end style close parentheses end fraction close parentheses therefore space space space space space space space space straight y equals tan to the power of negative 1 end exponent fraction numerator 3 straight x over denominator straight a end fraction plus tan to the power of negative 1 end exponent fraction numerator 2 straight x over denominator straight a end fraction space open square brackets because space tan to the power of negative 1 end exponent space straight A plus tan to the power of negative 1 end exponent straight B equals tan to the power of negative 1 end exponent space fraction numerator straight A plus straight B over denominator 1 minus AB end fraction close square brackets space therefore space dy over dx equals fraction numerator 1 over denominator 1 plus begin display style fraction numerator 9 straight x squared over denominator straight a squared end fraction end style end fraction. straight d over dx open parentheses fraction numerator 3 straight x over denominator straight a end fraction close parentheses plus fraction numerator 1 over denominator 1 plus begin display style fraction numerator 4 straight x squared over denominator straight a squared end fraction end style end fraction. straight d over dx open parentheses fraction numerator 2 straight x over denominator straight a end fraction close parentheses space space space space space space space space space space space space space space equals fraction numerator straight a squared over denominator straight a squared plus 9 space straight x squared end fraction.3 over straight a plus fraction numerator straight a squared over denominator straight a squared plus 4 space straight x squared end fraction.2 over straight a equals fraction numerator 3 straight a over denominator straight a squared plus 9 space straight x squared end fraction plus fraction numerator 2 straight a over denominator straight a squared plus 4 space straight x squared end fraction

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