How do you differentiate #f(x)=xln(1/x)-lnx/x #?

Question

How do you differentiate #f(x)=xln(1/x)-lnx/x   #?

William Abdalla · Accepted Answer

It is #f'(x)=ln(1/x)-1-(1-lnx)/x^2#

The derivative is

#d(f(x))/dx=x'ln(1/x)+xln(1/x)'-((lnx)'x-x'lnx)/(x^2)=> d(f(x))/dx=ln(1/x)+x((1/x)'/(1/x))-(1-lnx)/x^2=> d(f(x))/dx=ln(1/x)-1-(1-lnx)/x^2#

Remember that

#d(lnx)/dx=1/x#

Elijah Abram · Answer

undefined To differentiate the function $ f(x) = x\ln\left(\frac{1}{x}\right) - \frac{\ln x}{x} $, use the product rule and quotient rule as needed. Here are the steps:

1. Apply the product rule to the term $ x\ln\left(\frac{1}{x}\right) $ and the quotient rule to the term $ -\frac{\ln x}{x} $.
2. Differentiate each term separately.
3. Combine the derivatives to find the overall derivative of the function.

Let's differentiate step by step:

1. For the term $ x\ln\left(\frac{1}{x}\right) $:
   $$ f_1(x) = x\ln\left(\frac{1}{x}\right) $$
   Apply the product rule: $ u = x $ and $ v = \ln\left(\frac{1}{x}\right) $.
   $$ f_1'(x) = u'v + uv' $$
   $$ f_1'(x) = (1)\ln\left(\frac{1}{x}\right) + x\left(\frac{-1}{x^2}\right) $$
   $$ f_1'(x) = -\ln\left(\frac{1}{x}\right) - \frac{1}{x} $$

2. For the term $ -\frac{\ln x}{x} $:
   $$ f_2(x) = -\frac{\ln x}{x} $$
   Apply the quotient rule: $ u = -\ln x $ and $ v = x $.
   $$ f_2'(x) = \frac{u'v - uv'}{v^2} $$
   $$ f_2'(x) = \frac{(-\frac{1}{x})(x) - (-\ln x)(1)}{x^2} $$
   $$ f_2'(x) = -\frac{1}{x} + \frac{\ln x}{x^2} $$

Combine the derivatives:

$$ f'(x) = f_1'(x) + f_2'(x) $$
$$ f'(x) = (-\ln\left(\frac{1}{x}\right) - \frac{1}{x}) + (-\frac{1}{x} + \frac{\ln x}{x^2}) $$
$$ f'(x) = -\ln\left(\frac{1}{x}\right) - \frac{2}{x} + \frac{\ln x}{x^2} $$

So, the derivative of the function $ f(x) $ is:

$$ f'(x) = -\ln\left(\frac{1}{x}\right) - \frac{2}{x} + \frac{\ln x}{x^2} $$