# How do you find the first and second derivative of x^2 lnx?

##### 1 Answer
Aug 31, 2016

$\frac{d y}{d x} = 2 x \cdot l n \left(x\right) + x \text{ 'The first derivative '}$
$\frac{{d}^{2} y}{d {x}^{2}} = 2 \cdot l n \left(x\right) + 3 \text{ 'The second derivative '}$

#### Explanation:

$y = {x}^{2} \cdot l n \left(x\right)$

$u = {x}^{2} \text{ ; } u ' = 2 x$

$v = l n \left(x\right) \text{ ; } {v}^{'} = \frac{1}{x}$

$y = u \cdot v$

$\frac{d y}{d x} = {u}^{'} \cdot v + {v}^{'} \cdot u$

$\frac{d y}{d x} = 2 x \cdot l n \left(x\right) + \frac{1}{\cancel{x}} \cdot {\cancel{x}}^{2}$

$\frac{d y}{d x} = 2 x \cdot l n \left(x\right) + x \text{ 'The first derivative '}$

$a = 2 x \text{ ; } {a}^{'} = 2$
$b = l n \left(x\right) \text{ ; } {b}^{'} = \frac{1}{x}$

$\frac{{d}^{2} y}{d {x}^{2}} = {a}^{'} \cdot b + {b}^{'} \cdot a + 1$

$\frac{{d}^{2} y}{d {x}^{2}} = 2 \cdot l n \left(x\right) + \frac{1}{\cancel{x}} \cdot 2 \cancel{x} + 1$

$\frac{{d}^{2} y}{d {x}^{2}} = 2 \cdot l n \left(x\right) + 2 + 1$

$\frac{{d}^{2} y}{d {x}^{2}} = 2 \cdot l n \left(x\right) + 3 \text{ 'The second derivative '}$