differentiate y^2=x^3(x-2) and hence, show that for y>0, dy/dx = (2x-3) sqrt(x/x-2)
To differentiate y^2=x^3(x-2), first we can rewrite the equation as y=±√(x^3(x-2)) or y=±x√(x-2). Since we are given that y>0, we will consider the positive square root, y=x√(x-2).
Now, we can differentiate y=x√(x-2) using the product rule and chain rule:
dy/dx = (x)(1/2)(x-2)^(-1/2) + √(x-2)
dy/dx = (1/2)√(x-2) + √(x-2)
dy/dx = (3/2)√(x-2)
Finally, we can substitute y back into the equation to show that dy/dx = (2x-3)√(x/(x-2)) when y>0.