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

3 Answers
Jun 18, 2018

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)

Jun 18, 2018

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!

Jun 18, 2018

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