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

Dec 22, 2015

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

#### Explanation:

Quotient rule states that for a function $y = f \frac{x}{g} \left(x\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g} {\left(x\right)}^{2}$

Product rule states that for a function $y = f \left(x\right) \cdot g \left(x\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

Additionally, in this case we'll use the chain rule once, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

Thus,

The derivative of $f \left(x\right) = - \frac{\ln \left(\frac{1}{x}\right)}{x} + x \ln x$ proceeds as follows:

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{- \left(\frac{1}{\frac{1}{x}}\right) \left(- \frac{1}{x} ^ 2\right) \cdot x + \ln \left(\frac{1}{x}\right) \left(1\right)}{x} ^ 2 + \left(1\right) \cdot \ln \left(x\right) + x \left(\frac{1}{x}\right)$

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{\left(\frac{1}{\cancel{x}}\right) \cancel{x} + \ln \left(\frac{1}{x}\right)}{x} ^ 2 + \ln x + \cancel{x} \cdot \left(\frac{1}{\cancel{x}}\right)$

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{1 + \ln \left(\frac{1}{x}\right)}{x} ^ 2 + \ln x + 1$