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

Question

1071 views around the world

You can reuse this answer
Creative Commons License

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#