Question

If ${\log }_{10}\left(\frac{x^3-y^3}{x^3-y^3}\right)=2$, then show that $\frac{dy}{dx}=\frac{-99x^2}{101y^2}$.

Answer 1

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

$\Rightarrow \frac{x^3-y^3}{x^3+y^3}={10}^2=100$

$\Rightarrow x^3-y^3=100x^3+100y^3$

$\Rightarrow -101y^3=99x^3$

Differentiate both sides with respect to $x$

$-101\left(3y^2\right)\frac{dy}{dx}=99\left(3x^2\right)$

$\therefore \frac{dy}{dx}=\frac{-99x^2}{101y^2}$

Hence proved.