# How do you differentiate y=(x+5)(2x-3)(3x^2+4)?

Apr 24, 2018

$y ' = \left(2 x - 3\right) \left(3 {x}^{2} + 4\right) + 2 \left(x + 5\right) \left(3 {x}^{2} + 4\right) + 6 x \left(2 x - 3\right) \left(x + 5\right)$

$y ' = 24 {x}^{3} + 63 {x}^{2} - 74 x + 28$

#### Explanation:

If $y = u v w$, where $u$, $v$, and $w$ are all functions of $x$, then:
$y ' = u v w ' + u v ' w + u ' v w$ (This can be found by doing a chain rule with two functions substitued as one, i.e. making $u v = z$)

$u = x + 5$
$u ' = 1$

$v = 2 x - 3$
$v ' = 2$

$w = 3 {x}^{2} + 4$
$w ' = 6 x$

$y ' = \left(2 x - 3\right) \left(3 {x}^{2} + 4\right) + 2 \left(x + 5\right) \left(3 {x}^{2} + 4\right) + 6 x \left(2 x - 3\right) \left(x + 5\right)$

$y ' = 6 {x}^{3} + 8 x - 9 {x}^{2} - 12 + 6 {x}^{3} + 8 x + 30 {x}^{2} + 40 + 12 {x}^{3} + 60 {x}^{2} - 18 {x}^{2} - 90 x$

$y ' = 24 {x}^{3} + 63 {x}^{2} - 74 x + 28$

Apr 24, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = 24 {x}^{3} + 63 {x}^{2} - 74 x + 28$

#### Explanation:

$\text{expand the factors and differentiate using the "color(blue)"power rule}$

•color(white)(x)d/dx(ax^n)=nax^(n-1)

$y = \left(x + 5\right) \left(2 x - 3\right) \left(3 {x}^{2} + 4\right)$

$\textcolor{w h i t e}{y} = 6 {x}^{4} + 21 {x}^{3} - 37 {x}^{2} + 28 x - 60$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 24 {x}^{3} + 63 {x}^{2} - 74 x + 28$