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CBSE Class 12-science Answered

If x square root of 1 plus y end root space plus y square root of 1 plus x end root equals 0 comma s h o w space t h a t space fraction numerator d y over denominator d x end fraction equals space minus 1 over open parentheses x plus 1 close parentheses squared
Asked by sunil2791 | 23 May, 2017, 12:26: PM
answered-by-expert Expert Answer
begin mathsize 16px style Given colon straight x square root of 1 plus straight y end root space plus space straight y square root of 1 plus straight x end root space equals space 0 rightwards double arrow straight x square root of 1 plus straight y end root space equals negative space straight y square root of 1 plus straight x end root space Squaring space on space both space sides comma rightwards double arrow straight x squared left parenthesis 1 plus straight y right parenthesis space equals space straight y squared left parenthesis 1 plus straight x right parenthesis rightwards double arrow straight x squared space plus space straight x squared straight y space equals space straight y to the power of 2 space end exponent plus space straight y squared straight x rightwards double arrow straight x to the power of 2 space end exponent space minus space straight y squared space equals space xy squared space minus space straight x squared straight y rightwards double arrow straight x to the power of 2 space end exponent space minus space straight y squared space equals space xy left parenthesis straight y space minus space straight x right parenthesis rightwards double arrow left parenthesis straight x plus straight y right parenthesis left parenthesis straight x minus straight y right parenthesis space equals space xy left parenthesis straight y minus straight x right parenthesis rightwards double arrow left parenthesis straight x space plus space straight y right parenthesis space left parenthesis straight x space minus space straight y right parenthesis space equals space minus xy left parenthesis straight x space minus space straight y right parenthesis rightwards double arrow straight x space plus space straight y space equals space minus xy rightwards double arrow straight x space plus space xy space plus straight y space equals space 0 rightwards double arrow straight x space plus space straight y left parenthesis 1 plus straight x right parenthesis space equals space 0 rightwards double arrow straight y left parenthesis 1 plus straight x right parenthesis space equals space minus straight x rightwards double arrow straight y space equals space fraction numerator negative straight x over denominator left parenthesis 1 plus straight x right parenthesis end fraction Now comma space differentiating space straight y space straight w. straight r. straight t. space straight x comma rightwards double arrow dy over dx space equals negative space open square brackets fraction numerator left parenthesis 1 plus straight x right parenthesis left parenthesis 1 right parenthesis minus straight x left parenthesis 0 plus 1 right parenthesis over denominator left parenthesis 1 plus straight x right parenthesis squared end fraction close square brackets space space space space space space space space space space space space equals negative space open square brackets fraction numerator 1 space plus space straight x space minus space straight x over denominator left parenthesis 1 space plus space straight x right parenthesis squared end fraction close square brackets space space space space space space space space space space space space equals space minus fraction numerator 1 over denominator left parenthesis 1 space plus space straight x right parenthesis squared end fraction end style
Answered by Sneha shidid | 23 May, 2017, 03:24: PM

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