How do you find the derivative of y=3(x^2+1)(2x^2-1)(2x+3)?

Jun 18, 2018

$y ' = 6 \left(- 1 + 3 x + 3 {x}^{2} + 12 {x}^{3} + 10 {x}^{4}\right)$

Explanation:

After the formula

$\left(u v w\right) ' = u ' v w + u v ' w + u v w '$
we get

$y ' = 3 \cdot 2 x \cdot \left(2 {x}^{2} - 1\right) \left(2 x + 3\right) + 3 \left({x}^{2} + 1\right) 4 x \left(2 x + 3\right) + 3 \left({x}^{2} + 1\right) \left(2 {x}^{2} - 1\right) \cdot 2$
Expanding and combining like terms we obtain

$y ' = 6 \left(- 1 + 3 x + 3 {x}^{2} + 12 {x}^{3} + 10 {x}^{4}\right)$

Jun 18, 2018

${y}^{'} = 60 {x}^{4} + 72 {x}^{3} + 18 {x}^{2} + 18 x - 6$

Explanation:

Here we will be using the product rule:

${\left(a b c d\right)}^{'} = {a}^{'} b c d + a {b}^{'} c d + a b {c}^{'} d + a b c {d}^{'}$

$\setminus \therefore {y}^{'} = 0 + \left(3\right) \left(2 x\right) \left(2 {x}^{2} - 1\right) \left(2 x + 3\right) + \left(3\right) \left({x}^{2} + 1\right) \left(4 x\right) \left(2 x + 3\right) + \left(3\right) \left({x}^{2} + 1\right) \left(2 {x}^{2} - 1\right) \left(2\right)$

When the expression is simplified, it becomes:

${y}^{'} = 60 {x}^{4} + 72 {x}^{3} + 18 {x}^{2} + 18 x - 6$

Hope that makes sense!

Jun 18, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = 60 {x}^{4} + 72 {x}^{3} + 18 {x}^{2} + 18 x - 6$

Explanation:

$\text{differentiate each term using the "color(blue)"power rule}$

•color(white)(x)d/dx(ax^n)=nax^(n-1)

$\text{expanding the factors gives}$

$y = 12 {x}^{5} + 18 {x}^{4} + 6 {x}^{3} + 9 {x}^{2} - 6 x - 9$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 60 {x}^{4} + 72 {x}^{3} + 18 {x}^{2} + 18 x - 6$