To differentiate y = x^x, I need to use logarithmic differentiation since both the base and exponent contain x.
Given: y = x^x
Step 1: Take the natural logarithm of both sides. ln y = ln(x^x) ln y = x ln x
Step 2: Differentiate both sides with respect to x. d/dx(ln y) = d/dx(x ln x)
Step 3: Apply the chain rule on the left side and product rule on the right side. Left side: d/dx(ln y) = (1/y) · dy/dx
Right side: d/dx(x ln x) = x · d/dx(ln x) + ln x · d/dx(x) = x · (1/x) + ln x · 1 = 1 + ln x
Step 4: Set them equal. (1/y) · dy/dx = 1 + ln x
Step 5: Solve for dy/dx. dy/dx = y(1 + ln x)
Step 6: Substitute back y = x^x. dy/dx = x^x(1 + ln x)
Therefore: dy/dx = x^x(1 + ln x)
Note: This derivative is defined for x > 0, since we need x^x and ln x to be defined.
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