Question

How do you use the product rule to differentiate `(2x^2-5)^3(x^3+6x)^(4/3)`?

Answer 1

`d/dx[(2x^2-5)^3*(x^3+6x)^(4/3)]`
`=4(2x^2-5)^(2)(x^3+6x)^(1/3)(5x^4+17x^2-10)`

Explanation:

The formula `d/dx(uv)=ud/dx(v)+vd/dx(u)`

let `u=(2x^2-5)^3` and `v=(x^3+6x)^(4/3)`

Now use the formula

`d/dx((2x^2-5)^3*(x^3+6x)^(4/3))=(2x^2-5)^3d/dx((x^3+6x)^(4/3))+((x^3+6x)^(4/3))d/dx((2x^2-5)^3)`

`d/dx((2x^2-5)^3*(x^3+6x)^(4/3))=(2x^2-5)^3*(4/3)((x^3+6x)^(4/3-1))*d/dx(x^3+6x)+((x^3+6x)^(4/3))*3((2x^2-5)^(3-1))*d/dx(2x^2-5)`

`d/dx((2x^2-5)^3*(x^3+6x)^(4/3))=(2x^2-5)^3*(4/3)((x^3+6x)^(4/3-1))*(3x^2+6)+((x^3+6x)^(4/3))*3((2x^2-5)^(3-1))(4x-0)`

`d/dx((2x^2-5)^3*(x^3+6x)^(4/3))`
`=4(2x^2-5)^3*((x^3+6x)^(1/3))(x^2+2)+12x(x^3+6x)^(4/3)(2x^2-5)^(2)`

`d/dx((2x^2-5)^3*(x^3+6x)^(4/3))`
`=4(2x^2-5)^(2)(x^3+6x)^(1/3)[(2x^2-5)(x^2+2)+3x(x^3+6x)]`

`d/dx((2x^2-5)^3*(x^3+6x)^(4/3))`
`=4(2x^2-5)^(2)(x^3+6x)^(1/3)[2x^4-x^2-10+3x^4+18x^2]`

`d/dx((2x^2-5)^3*(x^3+6x)^(4/3))`
`=4(2x^2-5)^(2)(x^3+6x)^(1/3)(5x^4+17x^2-10)`

God bless....I hope the explanation is useful.