# How do you differentiate f(x)= x^2/ (lnx) twice using the quotient rule?

Feb 28, 2016

$f ' ' \left(x\right) = \frac{2 {\ln}^{2} x - 3 \ln x + 2}{\ln} ^ 3 x$

#### Explanation:

The quotient rule states that

$\frac{d}{\mathrm{dx}} \left(g \frac{x}{h \left(x\right)}\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$

So, for $f \left(x\right)$, we see that

$g \left(x\right) = {x}^{2}$
$h \left(x\right) = \ln x$

Differentiate both of these:

$g ' \left(x\right) = 2 x$
$h ' \left(x\right) = \frac{1}{x}$

Plugging these both in, we see that

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

Which simplifies to be

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

We can apply the quotient rule to this again to find the second derivative. We will redefine our functions being divided:

$g \left(x\right) = 2 x \ln x - x$
$h \left(x\right) = {\ln}^{2} x$

However, this time, finding $g ' \left(x\right)$ and $h ' \left(x\right)$ will not be so straightforward.

The $2 x \ln x$ term of $g \left(x\right)$ will require the product rule to differentiate.

We see that

$g ' \left(x\right) = \ln x \frac{d}{\mathrm{dx}} \left(2 x\right) + 2 x \frac{d}{\mathrm{dx}} \left(\ln x\right) - 1$

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

$g ' \left(x\right) = 2 \ln x + 1$

Finding $h ' \left(x\right)$ will require use of the chain rule:

$h ' \left(x\right) = 2 \ln x \frac{d}{\mathrm{dx}} \left(\ln x\right) = 2 \ln x \left(\frac{1}{x}\right) = \frac{2 \ln x}{x}$

Thus, we see that

$g \left(x\right) = 2 x \ln x - x$
$h \left(x\right) = {\ln}^{2} x$

$g ' \left(x\right) = 2 \ln x + 1$
$h ' \left(x\right) = \frac{2 \ln x}{x}$

Hence the second derivative equals

$f ' ' \left(x\right) = \frac{{\ln}^{2} x \left(2 \ln x + 1\right) - \left(2 x \ln x - x\right) \frac{2 \ln x}{x}}{\ln} ^ 4 x$

Expanding completely, this equals

$f ' ' \left(x\right) = \frac{2 {\ln}^{3} x + {\ln}^{2} x - 4 {\ln}^{2} x + 2 \ln x}{\ln} ^ 4 x$

Combining the ${\ln}^{2} x$ terms and dividing a common $\ln x$ term from the fraction, we see that

$f ' ' \left(x\right) = \frac{2 {\ln}^{2} x - 3 \ln x + 2}{\ln} ^ 3 x$