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)

$\frac{y}{x}$

Answer 2

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