If x ,e^ xy = y + ^2 x , then the value of dy dx at x = 0 is equal to

Step-by-step solution

Given equation is x e^ xy = y + ^2 x . Substituting x = 0 to find the corresponding value of y : 0 e^0 = y + ^2 0 y = 0 Differentiating the given equation with respect to x : d dx (x e^ xy ) = d dx (y + ^2 x) e^ xy + x e^ xy (y + x dy dx ) = dy dx + 2 x x Substituting x = 0 and y = 0 into the differentiated equation: e^0 + 0 e^0 (0 + 0 dy dx ) = dy dx + 2 0 0 1 + 0 = dy dx + 0 dy dx = 1 Answer: 1