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differentiate the function f(x)= xln(x)-x
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Use the Product rule: (uv)' = u'v+uv'
Of course the derivative of the "-x" part is just -1, so you have to use the Product Rule for xlnx
Where the derivative of ln(x)=1/x
Yeah, \(\large ln(x)=\dfrac{1}{x}\).
So we get: \(f'(x)=1 \cdot \ln x+x \cdot \dfrac{1}{x}-1\). What does this simplify to?
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