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If `x^(2) y^(5) = (x + y)^(7) , " then " (d^(2)y)/(dx^(2))` is equal toA. `y //x ^(2)`B. `x//y`C. 1D. 0 |
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Answer» Correct Answer - D Given , `x^(2) y^(2) = (x + y)^(7)` Taking log on both sides , we get 2 log x + 5 log y = 7 log ( x + y) On differentiating , we get `(2)/(x) + (5)/(y) (dy)/(dx) = (7 )/( x + y) = ( 1 + (dy)/(dx))` `implies (dy)/(dx) ((7)/(x+y) - (5)/(y)) = (2)/(x) - (7)/(x + y)` `implies (dy)/(dx) = (y)/(x) " " .... (i)` Again , differentiating , we get `(d^(2) y)/(dx^(2)) = ( x ""(dy)/(dx) - y)/(x^(2))` ` = (x * (y //x) - y)/(x^(2)) `[ from Eq. (i) ] = 0 |
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