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MHT CET · Maths · Differentiation

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

  1. A \(\left(-\frac{99}{101}\right) \frac{x^{2}}{y^{2}}\)
  2. B \(\left(-\frac{101}{99}\right) \frac{x^{2}}{y^{2}}\)
  3. C \(\left(-\frac{101}{99}\right) \frac{y^{2}}{x^{2}}\)
  4. D \(\left(-\frac{99}{101}\right) \frac{y^{2}}{x^{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left(-\frac{101}{99}\right) \frac{y^{2}}{x^{2}}\)

Step-by-step Solution

Detailed explanation

\(\text {Given } \log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2 \Rightarrow \frac{x^{3}-y^{3}}{x^{3}+y^{3}}=10^{2}\) \(\ldots\left[\because \log _{a} m=x \Rightarrow a^{x}=m\right] \)
\( \therefore x^{3}-y^{3}=100\left(x^{3}+y^{3}\right)\)
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 ]
\(\quad-99 x^{2}=101 y^{2} \frac{d y}{d x} \Rightarrow \frac{d x}{d y}=\left(-\frac{101}{99}\right) \frac{y^{2}}{x^{2}}\)