Find #dy/dx# if #y^2=x(x-1)^2#?
1 Answer
May 25, 2017
# dy/dx = ((x-1)(3x-1))/(2y) #
Explanation:
We have:
# y^2 = x(x-1)^2 #
Differentiating LHS implicitly, and RHS using the product rule along with the chain rule we get:
# (d/dyy^2)(dy/dx) = (x)(d/dx(x-1)^2) + (d/dx x)((x-1)^2) #
# :. 2ydy/dx = x(2(x-1)(1)) + (1)(x-1)^2 #
# :. 2ydy/dx = 2x(x-1) + (x-1)^2 #
# :. 2ydy/dx = (x-1)(2x+(x-1)) #
# :. 2ydy/dx = (x-1)(3x-1) #
# :. \ \ \ dy/dx = ((x-1)(3x-1))/(2y) #
