If $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$ then $\frac{d x}{d y}=$
If $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$ then $\frac{d x}{d y}=$
- $\left(-\frac{99}{101}\right) \frac{x^{2}}{y^{2}}$
- $\left(-\frac{101}{99}\right) \frac{x^{2}}{y^{2}}$
- $\left(-\frac{101}{99}\right) \frac{y^{2}}{x^{2}}$
- $\left(-\frac{99}{101}\right) \frac{y^{2}}{x^{2}}$
Solution
$\begin{array}{l}
\text { Given } \log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2 \Rightarrow \frac{x^{3}-y^{3}}{x^{3}+y^{3}}=10^{2} \quad \ldots\left[\because \log _{a} m=x \Rightarrow a^{x}=m\right] \\
\therefore x^{3}-y^{3}=100\left(x^{3}+y^{3}\right)
\end{array}$
Differentiating w.r.t. $x$,
$\begin{array}{l}
\quad 3 x^{2}-3 y^{2} \frac{d y}{d x}=100\left(3 x^{2}+3 y^{2} \frac{d y}{d x}\right)\end{array}$
$\therefore x^{2}-100 x^{2}=y^{2}(1+100) \frac{d y}{d x}$ ...[Dividing both sides by 3 ]
$\quad-99 x^{2}=101 y^{2} \frac{d y}{d x} \Rightarrow \frac{d x}{d y}=\left(-\frac{101}{99}\right) \frac{y^{2}}{x^{2}}$
Asked in: MHT CET 2020 (13 Oct Shift 1)
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