How do you find the second derivative of f(x)=x^2 lnx ?

2 Answers
Jun 1, 2018

$f ' ' \left(x\right) = 2 \ln \left(x\right) + 3$

Explanation:

By the product rule we get
$f ' \left(x\right) = 2 x \ln \left(x\right) + {x}^{2} \cdot 17 x$
simplifying
$f ' \left(x\right) = 2 x \ln \left(x\right) + x$
$f ' ' \left(x\right) = 2 \ln \left(x\right) + 2 x \cdot \frac{1}{x} + 1$
simplifying we get
$f ' ' \left(x\right) = 2 \ln \left(x\right) + 3$

Jun 1, 2018

$f ' ' \left(x\right) = 3 + 2 \ln x$

Explanation:

$\text{differentiate using the "color(blue)"product rule}$

$\text{given "f(x)=g(x)h(x)" then}$

$f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$

$g \left(x\right) = {x}^{2} \Rightarrow g ' \left(x\right) = 2 x$

$h \left(x\right) = \ln x \Rightarrow h ' \left(x\right) = \frac{1}{x}$

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

$\text{differentiate "2xlnx" using the "color(blue)"product rule}$

$g \left(x\right) = 2 x \Rightarrow g ' \left(x\right) = 2$

$h \left(x\right) = \ln x \Rightarrow h ' \left(x\right) = \frac{1}{x}$

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

$f ' ' \left(x\right) = 1 + 2 + 2 \ln x = 3 + 2 \ln x$