Question

How do you find the derivative of `(3x+1)^(3/2) (2x+4)`?

Answer 1

`2(3x+1)^(3/2)+(9(x+2)(3x+1)^(1/2))`

Explanation:

it can be found by uv rule
let us assume `u=(3x+1)^(3/2)` and `v=2x+4`
from basic formula of diferentiation
`d(uv)/dx`=`ud/dxv+vd/dxu`=`uv'+vu'`
`u=(3x+1)^(3/2) v=2x+4`
`u'=(3/2)(3x+1)^(3/2-1)3`
`=>u'=9/2(3x+1)^(1/2)`
`v'=2`
substituiting all the values
`d/dx(3x+1)^(3/2)(2x+4)`=`2(3x+1)^(3/2)+((2x+4)9/2(3x+1)^(1/2))`
=`2(3x+1)^(3/2)+(9(x+2)(3x+1)^(1/2))`