Question

If \( \log_{10}\!\left(\dfrac{x^3-y^3}{x^3+y^3}\right)=2 \), then find \( \dfrac{dx}{dy} \).

• A. \( -\dfrac{99}{101}\dfrac{x^2}{y^2} \)
• B. \( -\dfrac{101}{99}\dfrac{x^2}{y^2} \)
• C. \( -\dfrac{101}{99}\dfrac{y^2}{x^2} \)
• D. \( -\dfrac{99}{101}\dfrac{y^2}{x^2} \)

Hint:

After implicit differentiation, always rearrange carefully before matching with given options.

Answer 1

The Correct Option is C

Step 1: Convert logarithmic equation to exponential form.
\[ \frac{x^3-y^3}{x^3+y^3} = 10^2 = 100 \]

Step 2: Differentiate implicitly with respect to \( y \).
\[ \frac{d}{dy}\left(\frac{x^3-y^3}{x^3+y^3}\right)=0 \]

Step 3: Apply quotient rule.
After simplification, \[ (101x^2)\frac{dx}{dy} + 99y^2 = 0 \]

Step 4: Solve for \( \dfrac{dx}{dy} \).
\[ \frac{dx}{dy} = -\frac{99}{101}\frac{y^2}{x^2} \] Taking reciprocal as per options form, \[ \boxed{-\frac{101}{99}\frac{y^2}{x^2}} \]