Question

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

• A. $\frac{x}{y}$
• B. $-\frac{y}{x}$
• C. $-\frac{x}{y}$
• D. $\frac{y}{x}$

Answer 1

Answer: (D)

$\frac{y}{x}$

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

$\frac{x^3-y^3}{x^3+y^3}=100$,

applying componendo and dividendo gives,

$\Rightarrow -\frac{x^3}{y^3}=\frac{101}{99}$,

$\Rightarrow \frac{x^3}{y^3}=-\frac{101}{99}$,

Differentiating with respect to $x$ gives,

$⟹3\frac{x^2}{y^2}\left(\frac{y-x\frac{dy}{dx}}{y^2}\right)=0$

$\therefore \frac{dy}{dx}=\frac{y}{x}$