If log _{10}left(frac{x^{3}-y^{3}}{x^{3}+y^{3 | vedclass

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(C) Given $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$.By the definition of logarithm,$\frac{x^{3}-y^{3}}{x^{3}+y^{3}}=10^{2}=100$.There

Vedclass · Accepted Answer

$\left(-\frac{101}{99}\right) \frac{y^{2}}{x^{2}}$

Vedclass · Answer

(C) Given $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$.By the definition of logarithm,$\frac{x^{3}-y^{3}}{x^{3}+y^{3}}=10^{2}=100$.Therefore,$x^{3}-y^{3}=100(x^{3}+y^{3})$.$x^{3}-y^{3}=100x^{3}+100y^{3}$.$-99x^{3}=101y^{3}$.Differentiating both sides with respect to $y$:$-99 \cdot 3x^{2} \frac{dx}{dy} = 101 \cdot 3y^{2}$.$-297x^{2} \frac{dx}{dy} = 303y^{2}$.$\frac{dx}{dy} = -\frac{303y^{2}}{297x^{2}} = -\frac{101y^{2}}{99x^{2}}$.Thus,$\frac{dx}{dy} = \left(-\frac{101}{99}\right) \frac{y^{2}}{x^{2}}$.

If $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$,then $\frac{dx}{dy} = $

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