Use logarithmic differentiation to find derivative of y=x^x^2 x > 0. Please show all steps so I can follow the steps. Thank you.!

y = x^x^2

lny = x^2 ln x
1/y y' = 2x lnx + x^2/x
= 2x lnx + x

y' = xy(2lnx + 1)
y' = x x^x^2 (2lnx + 1)

To find the derivative of the function y = x^(x^2), we can use logarithmic differentiation. Here are the steps:

Step 1: Take the natural logarithm of both sides of the equation:
ln(y) = ln(x^(x^2))

Step 2: Use logarithm properties to simplify the right-hand side. Apply the power rule of logarithms:
ln(y) = (x^2) ln(x)

Step 3: Differentiate implicitly with respect to x on both sides of the equation. The derivative of ln(y) with respect to x can be expressed as (dy/dx) / y using the chain rule, and the derivative of (x^2) ln(x) is found using the product rule:
(1/y) (dy/dx) = (2x) ln(x) + (x^2)(1/x)

Step 4: Solve for (dy/dx), the derivative of y with respect to x. Multiply both sides of the equation by y:
dy/dx = y * [(2x) ln(x) + (x^2)(1/x)]

Step 5: Replace y with its original expression, x^(x^2):
dy/dx = x^(x^2) * [(2x) ln(x) + (x^2)(1/x)]

And there you have it. The derivative of y = x^(x^2) with respect to x is given by:

dy/dx = x^(x^2) * [(2x) ln(x) + (x^2)(1/x)]