If _ 10 (x^2-y^2 x^2+y^2 )=2, then d y d x = - Vidyadip

MCQ

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

✓
$-\frac{99 x}{101 y}$
B
$\frac{99 x}{101 y}$
C
$-\frac{99 y}{101 x}$
D
$\frac{99 y}{101 x}$

✓

Answer

Correct option: A.

$-\frac{99 x}{101 y}$

$\log _{10}\left(\frac{x^2-y^2}{x^2+y^2}\right)=2\ ($Given$)$
$\Rightarrow \frac{x^2-y^2}{x^2+y^2}=100$
$\Rightarrow x^2-y^2=100 x^2+100 y^2$
$ \Rightarrow 99 x^2+101 y^2=0$
Differentiate $\text{w.r.t}. \ x$, we get
$2(99) x+2(101) y \frac{d y}{d x}=0$
$\Rightarrow 99 x+101 y \frac{d y}{d x}=0 $
$\Rightarrow \frac{d y}{d x}=-\frac{99 x}{101 y}$

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