What is the derivative of $f(x)=-ln(1/x)/x+xlnx$ ?

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Answer 1 · 2015-12-22T21:03:57

For the first term, we use quotient rule and for the second, product rule.

Quotient rule states that for a function $y=f(x)/g(x)$ , $(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/g(x)^2$

Product rule states that for a function $y=f(x)*g(x)$ , $(dy)/(dx)=f'(x)g(x)+f(x)g'(x)$

Additionally, in this case we'll use the chain rule once, which states that $(dy)/(dx)=(dy)/(du)(du)/(dx)$

Thus,

The derivative of $f(x)=-(ln(1/x))/x+xlnx$ proceeds as follows:

$(df(x))/(dx)=(-(1/(1/x))(-1/x^2)*x+ln(1/x)(1))/x^2 + (1)*ln(x)+x(1/x)$

$(df(x))/(dx)=((1/cancel(x))cancel(x)+ln(1/x))/x^2+lnx+cancel(x)*(1/cancel(x))$

$(df(x))/(dx)=(1+ln(1/x))/x^2+lnx+1$

What is the derivative of f(x)=-ln(1/x)/x+xlnxf(x)=-ln(1/x)/x+xlnx?