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Let f(x)=x^x for x>0, then find f'(x) for all x>0.

A common misconception from many students when tackling this problem is that they think the usual 'power rule' works. However, in this case the power is itself a function of x and not just a constant, so this would not work. To solve this problem, we will have to 'get rid' of the power. We will do this using the natural logarithm. ln(f(x))=xln(x) (1)Differentiating (1) and using the product rule on the right hand side and the chain rule on the left hand side, we get f'(x)/f(x)=ln(x)+1 Lastly rearranging for f'(x) and substituting for f(x), we derived f'(x)=x^x ( ln(x)+1) as required. This technique is known as logarithmic differentiation.

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Let f(x)=x^x for x>0, then find f'(x) for all x>0.

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