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)

Given,
${\log }_{10}\left(\frac{x^3-y^3}{x^3+y^3}\right)=2$
$\Rightarrow \frac{x^3-y^3}{x^3+y^3}=10^2=100$
$\Rightarrow x^3-y^3=100\left(x^3+y^3\right)$
$\Rightarrow 101y^3=-99x^3$
On differentiating both sides w.r.t. $x$ we get
$101\times 3y^2\frac{dy}{dx}=-99\cdot \left(3x^2\right)$
$\Rightarrow 101y^2\frac{dy}{dx}=-99x^2$
On multiplying by $x$ both sides, we get
$\Rightarrow 101x{y}^2\frac{dy}{dx}=-99x^3$
$\Rightarrow \frac{dy}{dx}=\frac{-99x^3}{101x{y}^2}$
$\Rightarrow \frac{dy}{dx}=\frac{101y^3}{101x{y}^2}\left[\because -99x^3=101y^3\right]$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$