y = x^x, find y'
y = xx, by taking logs:ln(y) = ln(xx )By log laws:ln(y) = xln(x)From implicit differentiation; d/dx = y'*d/dy, so:d/dx[ln(y)] = y'*d/dy[ln(y)] = y' * 1/yFrom the product rule; d/dx[xln(x)] = ln(x) * d/dx(x) + x * d/dx[ln(x)], so:d/dx[xln(x)] = ln(x) + 1Since d/dx[ln(y)] = d/dx[xln(x)] then:y' * 1/y = ln(x) + 1, y' = y[ln(x) + 1]y = xxFinally: y' = xx [ln(x) + 1]
SD
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