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If ` log _(10)((x^(3)-y^(3))/(x^(3)+y^(3))) = 2`, then show that ` (dy)/(dx) = (-99x^(2))/(101y^(2))` |
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Answer» Given, `log_(10)(x^(3)-y^(3))/(x^(3)+y^(3)) = 2` Or `log_(10)(x^(3)-y^(3)) - log_(10)(x^(3)+y^(3))=2` Now, differentiating with respect to x, we have , ` 1/(x^(3)-y^(3)) (3x^(2)-3y^(2)(dy)/(dx)) - 1/(x^(3)+y^(3))(3x^(2)+3y^(2)(dy)/(dx))` or `(3x^(2))/(x^(3)-y^(3))-(3y^(2))/(x^(3)-y^(3))(dy)/(dx)-(3x^(2))/(x^(3)+y^(3)) - (3y^(2))/(x^(3)+y^(3))(dy)/(dx)=0` or ` - (3y^(2))/(x^(3)-y^(3))(dy)/(dx)-(3y^(2))/(x^(3)+y^(3))(dy)/(dx) = - (3x^(2))/(x^(3)-y^(3))+(3x^(2))/(x^(3)+y^(3))` or ` -3y^(2)(dy)/(dx)[1/(x^(3)-y^(3))+1/(x^(3)+y^(3))]` `=-3x^(2)[1/(x^(3)-y^(3))-1/(x^(3)+y^(3))]` or `y^(2)(dy)/(dx) [1/(x^(3)-y^(3))-1/(x^(3)+y^(3))]` ` = x^(2)[1/(x^(3)-y^(3))-1/(x^(3)+y^(3))]` or `y^(2)(dy)/(dx)[(x^(3)+y^(3)+x^(3)-y^(3))/((x^(3)-y^(3))(x^(3)+y^(3)))]` ` = x^(2)[(x^(3)+y^(3)-x^(3)+y^(3))/((x^(3)-y^(3))(x^(3)+y^(3)))]` or ` y^(2)(dy)/(dx) [(2x^(3))/((x^(3)-y^(3))(x^(3)+y^(3)))]` ` = x^(2)[(2y^(3))/((x^(3)-y^(3))(x^(3)+y^(3)))]` or ` x(dy)/(dx) = y` or ` (dy)/(dx) = y/ x ` ... (i) Also, ` log_(10) . (x^(3)-y^(3))/(x^(3)+y^(3)) = 2 ` or ` (x^(3)-y^(3))/(x^(3)+y^(3)) = 10^(2)` or `(x^(3)-y^(3))/(x^(3)+y^(3)) = 100` or ` x^(3) - y^(3) = 100 (x^(3)+y^(3))` or ` x^(3)-y^(3) = 100 x^(3) + 100 y^(3)` or ` -99 x^(3) = 101 y^(3)` or ` y^(3)/x^(3) = - 99/101` or ` y/x= - (99x^(2))/(101 y^(2))` ...(ii) From (i) and (ii) , we get, ` (dy)/(dx) = - (99x^(2))/(101 y^(2))` |
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