Question

Find y' by implicit differentiation √x +√y =1

Answer 1

Differentiate both sides with respect to x


`(d)/(dx)(x^{(1)/(2)} + y^{(1)/(2)}) = (d)/(dx)(1)\\\\ \frac{1}{2x^{(1)/(2)}} + \frac{1}{2y^{(1)/(2)}}(dy)/(dx) = 0\\\\\frac{1}{2y^{(1)/(2)}}(dy)/(dx) = -\frac{1}{2x^{(1)/(2)}}\\\\(1)/(2√(y))(dy)/(dx) = -(1)/(2√(x)) \\\\(dy)/(dx) = -(1)/(2√(x)) * 2√(y)\\\\(dy)/(dx) = -(√(y))/(√(x)) \\\\\boxed{\bf{y'= -(√(y))/(√(x))}}`