Let f(x)=xx f(x) = x^x f(x)=xx.
Take the natural logarithm of both sides: ln(f(x))=ln(xx)=xln(x) \ln(f(x)) = \ln(x^x) = x \ln(x) ln(f(x))=ln(xx)=xln(x)
Differentiate both sides with respect to x x x: ddx(ln(f(x)))=ddx(xln(x)) \frac{d}{dx}(\ln(f(x))) = \frac{d}{dx}(x \ln(x)) dxd(ln(f(x)))=dxd(xln(x))
Using the chain rule on the left side and the product rule on the right side, we get: 1f(x)⋅f′(x)=ln(x)+1 \frac{1}{f(x)} \cdot f'(x) = \ln(x) + 1 f(x)1⋅f′(x)=ln(x)+1
Multiply both sides by f(x) f(x) f(x) to isolate f′(x) f'(x) f′(x): f′(x)=f(x)(ln(x)+1) f'(x) = f(x)(\ln(x) + 1) f′(x)=f(x)(ln(x)+1)
Substituting back f(x)=xx f(x) = x^x f(x)=xx: f′(x)=xx(ln(x)+1) f'(x) = x^x(\ln(x) + 1) f′(x)=xx(ln(x)+1)
f′(x)=xx(ln(x)+1)\boxed{f'(x) = x^x(\ln(x) + 1)}f′(x)=xx(ln(x)+1)
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