If y=Root of(logx+root of (logx+root of logx.........infinty)...prove that dy/dx=1/x(2y-1)
y = √(lnx+√(lnx+√(lnx+...)))
y^2 = lnx+√(lnx+√(lnx+...))) = lnx + y
2y y' = 1/x + y'
(2y-1)y' = 1/x
y' = 1/(x(2y-1))
To prove that dy/dx = 1/x(2y - 1), we need to take the derivative of both sides of the equation y = √(logx + √(logx + √(logx + ...))), and then solve for dy/dx.
Let's take the derivative step by step.
Step 1: Let's differentiate y with respect to x.
dy/dx = d/dx (√(logx + √(logx + √(logx + ...))))
Step 2: We can rewrite the right-hand side using the recursive definition of y:
y = √(logx + y)
Step 3: Square both sides of the equation to eliminate the square root:
y^2 = logx + y
Step 4: Rearrange the equation to solve for y in terms of logx:
y^2 - y = logx
Step 5: Now, differentiate both sides of the equation with respect to x using the chain rule. Let's find d(y^2 - y)/dx first:
d(y^2 - y)/dx = d(logx)/dx
Step 6: Differentiate the left-hand side using the chain rule. Applying the power rule, we have:
2y * dy/dx - dy/dx = 1/x
Step 7: Combine like terms:
(2y - 1) * dy/dx = 1/x
Step 8: Solve for dy/dx:
dy/dx = 1/(x * (2y - 1))
Therefore, we have proven that dy/dx = 1/(x * (2y - 1)).