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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation2 Marks
MCQ
If $\log 10\left(\frac{x^3-y^3}{x^3+y^3}\right)=2$, then $\frac{d y}{d x}=$
  • A
    $\frac{x}{y}$
  • B
    $-\frac{y}{x}$
  • C
    $-\frac{x}{y}$
  • ✓
    $\frac{y}{x}$
✓

Answer

Correct option: D.
$\frac{y}{x}$
We have, $\log _{10}\left(\frac{x^3-y^3}{x^3+y^3}\right)=2$
$\Rightarrow \log \left(x^3-y^3\right)-\log \left(x^3+y^3\right)=2 \log 10$
On differentiating $\text{w.r.t.} \ x$, we get
$\frac{1}{x^3-y^3}\left(3 x^2-3 y^2 \frac{d y}{d x}\right)-\frac{1}{x^3+y^3}\left(3 x^2+3 y^2 \frac{d y}{d x}\right)=0$
$\Rightarrow \frac{3 x^2}{x^3-y^3}-\frac{3 x^2}{x^3+y^3}-\left(\frac{3 y^2}{x^3-y^3}+\frac{3 y^2}{x^3+y^3}\right) \frac{d y}{d x}=0$
$\Rightarrow \left(3 y^2 \times 2 x^3\right) \frac{d y}{d x}$
$=3 x^2 \times 2 y^3 $
$\Rightarrow \frac{d y}{d x}=\frac{y}{x}$

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