Find the derivative of y = x^x
Hint: Write x^x as exp[x log(x)]
if u and v are functions of x, and
y = u^v
y' = vu^(v-1) u' + lnv u^v v'
= u^v (v/u u' + lnv v')
sort of a combination of the power rule and the exponent rule.
Count Iblis's suggestion is also useful, and can be done by noting that
ln y = x logx
1/y y' = 1 + logx
To find the derivative of y = x^x, we can use logarithmic differentiation.
Step 1: Take the natural logarithm (ln) of both sides of the equation:
ln(y) = ln(x^x)
Step 2: Use the properties of logarithms to simplify the expression:
ln(y) = x ln(x)
Step 3: Differentiate implicitly with respect to x:
(1/y) * (dy/dx) = ln(x) + x * (1/x)
Step 4: Simplify the expression:
dy/dx = y * (ln(x) + 1)
Step 5: Substitute y = x^x back into the equation:
dy/dx = x^x * (ln(x) + 1)
Therefore, the derivative of y = x^x is dy/dx = x^x * (ln(x) + 1).