Question

if log (x sq -y sq /x sq + y sq ) =a , then dy/dx =

Answer 1

Answer:

$dy/dx$ = {2x + 2 ($y^{2}/x^3$)} / {2y - 2($y/x^{2}$)}

Step-by-step explanation:

Given log($x^{2} -y^{2}/x^{2} +y^{2}$) = a

       ⇒ $x^{2} -y^{2}/x^{2} +y^{2}$ = $e^a$

       Differentiating both sides with respect to x we get,

                2x + 2($y^{2}/x^3$) - 2($y/x^{2}$) $dy/dx$ + 2y $dy/dx$ = 0

             ⇒$dy/dx$ = {2x + 2 ($y^{2}/x^3$)} / {2y - 2($y/x^{2}$)}