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

Question

William Abdalla · Accepted Answer

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

Paisley Johnson · Answer

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

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)#

Christopher Allers · Answer

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

The product rule will be applied in this case:

#(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)#

The expression becomes: after it is simplified.

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

I hope that was clear.

Ezra Adams · Answer

undefined To find the derivative of $ y = 3(x^2+1)(2x^2-1)(2x+3) $, you would apply the product rule, which states that the derivative of a product of functions is the sum of the derivative of each function times the other functions. Using the product rule, the derivative of the given function is:

$$ y' = 3[(x^2+1)(4x(2x^2-1)(2x+3)) + (x^2+1)(2x^2-1)(4x(2x+3))] $$

Simplify the expression:

$$ y' = 3[(4x^3+4x)(2x^2-1)(2x+3) + (2x^2-1)(4x)(2x+3)] $$

$$ y' = 3[(8x^5+12x^4-4x^3+8x^3-12x^2+4x)+(8x^3-4x^2+12x^2-6x)] $$

$$ y' = 3[8x^5+12x^4+4x-12x^2+8x^3-12x^2+8x^3-4x^2+12x^2-6x] $$

$$ y' = 3[8x^5+24x^3-16x^2-6x] $$

$$ y' = 24x^3+72x^3-48x^2-18x $$