Question

$Find space dy over dx space of space the space following space cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses comma space 0 less than straight x less than 1 space straight w. straight r. straight t. straight x. space$

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Solution

Let space space space space straight y equals cos to the power of negative 1 end exponent fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction Put space space space space straight x equals tan space straight theta therefore space space space space space space straight y equals cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses equals cos to the power of negative 1 end exponent left parenthesis cos space 2 straight theta right parenthesis equals 2 straight theta therefore space space space space space space straight y equals 2 space tan to the power of negative 1 end exponent straight x therefore space dy over dx equals fraction numerator 2 over denominator 1 plus straight x squared end fraction.

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