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Asked by sarveshvibrantacademy | 23 May, 2019, 00:08: AM
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L e t space u equals cos to the power of negative 1 end exponent square root of x space a n d space v equals square root of 1 minus x end root rightwards double arrow fraction numerator d u over denominator d x end fraction equals fraction numerator negative 1 over denominator square root of 1 minus x squared end root end fraction space a n d space fraction numerator d v over denominator d x end fraction equals fraction numerator negative 1 over denominator 2 square root of 1 minus x end root end fraction N o w comma fraction numerator d open parentheses cos to the power of negative 1 end exponent square root of x close parentheses over denominator d open parentheses square root of 1 minus x end root close parentheses end fraction equals fraction numerator d u over denominator d v end fraction equals fraction numerator begin display style fraction numerator d u over denominator d x end fraction end style over denominator begin display style fraction numerator d v over denominator d x end fraction end style end fraction equals fraction numerator fraction numerator negative 1 over denominator square root of 1 minus x squared end root end fraction over denominator fraction numerator negative 1 over denominator 2 square root of 1 minus x end root end fraction end fraction equals fraction numerator 2 square root of 1 minus x end root over denominator square root of 1 minus x squared end root end fraction equals fraction numerator 2 square root of 1 minus x end root over denominator square root of left parenthesis 1 minus x right parenthesis left parenthesis 1 plus x right parenthesis end root end fraction equals fraction numerator 2 over denominator square root of 1 plus x end root end fraction
Answered by Renu Varma | 23 May, 2019, 11:26: AM

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