Using the chain rule, the derivative of f(g(x)) is f'(g(x)) . g'(x).
If f(x) = ln(x), then f'(x) = 1/x, right?
Likewise, if g(x) = ln(x), then g'(x) = 1/x.
Then, f(g(x)) is ln(ln(x)). The derivative of this is f'(g(x)).g'(x) = f'(ln(x)) . 1/x = 1/ln(x) . 1/x = 1/xlnx.
For 2), you can use the product rule to differentiate x ln(x). When you do this, you get 1+ln(x). Therefore, the derivative of xln(x)-x = 1+ln(x)-1 = ln(x).
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