Question

How do you differentiate `y=(x+5)(2x-3)(3x^2+4)`?

Answer 1

`y'=(2x-3)(3x^2+4)+2(x+5)(3x^2+4)+6x(2x-3)(x+5)`

`y'=24x^3+63x^2-74x+28`

Explanation:

If `y=uvw`, where `u`, `v`, and `w` are all functions of `x`, then:
`y'=uvw'+uv'w+u'vw` (This can be found by doing a chain rule with two functions substitued as one, i.e. making `uv=z`)

`u=x+5`
`u'=1`

`v=2x-3`
`v'=2`

`w=3x^2+4`
`w'=6x`

`y'=(2x-3)(3x^2+4)+2(x+5)(3x^2+4)+6x(2x-3)(x+5)`

`y'=6x^3+8x-9x^2-12+6x^3+8x+30x^2+40+12x^3+60x^2-18x^2-90x`

`y'=24x^3+63x^2-74x+28`

Answer 2

`dy/dx=24x^3+63x^2-74x+28`

Explanation:

`"expand the factors and differentiate using the "color(blue)"power rule"`

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

`y=(x+5)(2x-3)(3x^2+4)`

`color(white)(y)=6x^4+21x^3-37x^2+28x-60`

`rArrdy/dx=24x^3+63x^2-74x+28`