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

3 Answers
Jun 18, 2018

Answer:

#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

Answer:

#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

Answer:

#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#