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If `y = "log "((1 + sqrtx)/(1 - sqrtx)) " then " (dy)/(dx) =` ?A. `(1)/(sqrtx (1-x))`B. `(-1)/(x(1 - sqrtx)^(2))`C. `(-sqrtx)/(2(1 - sqrtx))`D. none of these

Answer» Correct Answer - A
`y = log (1 + sqrtx) - log (1 - sqrtx)`
`rArr (dy)/(dx) = {(1)/(2 sqrtx (1 + sqrtx)) + (1)/(2 sqrtx (1 - sqrtx))} = (1)/(2 sqrtx) .(2)/((1 -x)) = (1)/(sqrtx (1 -x))`


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