Question

If ${\log }_{10}\frac{x^3-y^3}{x^3+y^3}=2$ then show that $\frac{dy}{dx}=\frac{-99x^2}{101y^2}$

Answer 1

$\begin{array}{c}{\log }_{10}\frac{\left(x^3-y^3\right)}{\left(x^3+y^3\right)=2}\\ \Rightarrow {10}^2=\frac{x^3-y^3}{x^3+y^3}\\ \Rightarrow 100x^3+100y^3-x^3+y^3=0\\ \Rightarrow 99x^3+101y^3=0\\ differentiatesw.r.to.xbothsides\\ \Rightarrow 101\times 3y^{2}dy/dx=-99\times 3x^2\\ \frac{dy}{dx}=\frac{-99x^2}{101y^2}\end{array}$