Question

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

Answer 1

`y'=6(-1+3x+3x^2+12x^3+10x^4)`

Explanation:

After the formula

`(uvw)'=u'vw+uv'w+uvw'`
we get

`y'=3*2x*(2x^2-1)(2x+3)+3(x^2+1)4x(2x+3)+3(x^2+1)(2x^2-1)*2`
Expanding and combining like terms we obtain

`y'=6(-1+3x+3x^2+12x^3+10x^4)`

Answer 2

`y^' = 60x^4+72x^3+18x^2+18x-6`

Explanation:

Here we will be using the product rule:

`(abcd)^' = a^'bcd + ab^'cd + abc^'d + abcd^'`

`\thereforey^' = 0 + (3)(2x)(2x^2-1)(2x+3) + (3)(x^2+1)(4x)(2x+3) + (3)(x^2+1)(2x^2-1)(2)`

When the expression is simplified, it becomes:

`y^' = 60x^4+72x^3+18x^2+18x-6`

Hope that makes sense!

Answer 3

`dy/dx=60x^4+72x^3+18x^2+18x-6`

Explanation:

`"differentiate each term using the "color(blue)"power rule"`

`•color(white)(x)d/dx(ax^n)=nax^(n-1)`

`"expanding the factors gives"`

`y=12x^5+18x^4+6x^3+9x^2-6x-9`

`dy/dx=60x^4+72x^3+18x^2+18x-6`