Question

Differentiate the following functions by substitutions method : $tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator square root of straight a squared minus straight x squared end root end fraction close parentheses$

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Solution

Let space space space straight y equals tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator square root of straight a squared minus straight x squared end root end fraction close parentheses Put space space space straight x equals straight a space sin space straight theta space therefore space space space space space straight y equals tan to the power of negative 1 end exponent open parentheses fraction numerator straight a space sinθ over denominator square root of straight a squared minus straight a squared sin squared end root straight theta end fraction close parentheses tan to the power of negative 1 end exponent open parentheses fraction numerator straight a space sinθ over denominator straight a space cos space straight theta end fraction close parentheses equals tan to the power of negative 1 end exponent left parenthesis tan space straight theta right parenthesis equals straight theta therefore space space space space space straight y equals sin to the power of negative 1 end exponent open parentheses straight x over straight a close parentheses therefore space dy over dx equals fraction numerator 1 over denominator square root of 1 minus begin display style straight x squared over straight a squared end style end root end fraction

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