# How do you find the derivative of f(x) = x²(3x³-1) and the given point is (1,2)?

Feb 21, 2018

13

#### Explanation:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} f \left(x\right)$

$= \frac{d}{\mathrm{dx}} {x}^{2} \left(3 {x}^{3} - 1\right)$ color(white)(dwwwwwdd $\left[\text{applying product rule}\right]$

$= \left(\frac{d}{\mathrm{dx}} {x}^{2}\right) \left(3 {x}^{3} - 1\right) + {x}^{2} \left(\frac{d}{\mathrm{dx}} 3 {x}^{3} - \frac{d}{\mathrm{dx}} 1\right)$

$= 2 x \left(3 {x}^{3} - 1\right) + {x}^{2} \left(9 {x}^{2}\right)$

$= 6 {x}^{4} - 2 x + 9 {x}^{4}$

$= 15 {x}^{4} - 2 x$

So, Derivative of f(x) at (1, 2) = $f ' \left(1\right) = 15 {\left(1\right)}^{4} - 2 \cdot 1 = 15 - 2 = 13$ color(white)(dwww(just replace $x = 1$)

Hence Explained.