Question

How do you differentiate `(3x-2)/(2x+1)^(1/2)`?

Answer 1

`(3x+5)/((2x+1)*sqrt(2x+1))`

Explanation:

By the Quotient rule we get

`(3sqrt(2x+1)-(3x-1)*(1/2)*(2x+1)^(-1/2)*2)/(2x+1)`
multiplying numerator and denominator by `sqrt(2x+1)`

we get

`((3(2x+1)-(3x-2)))/((2x+1)*sqrt(2x+1))`
simplifying we get

`(3x+5)/((2x+1)sqrt(2x+1))`

Answer 2

`f'(x)=(3x+1)/(2x+1)^(3/2)`

Explanation:

With the quotient rule.

`f(x)=(u(x))/(v(x))`
`f'(x)=(u'(x)v(x)-u(x)v'(x))/(v(x)^2)`
`f(x)=(3x+2)/((2x+1)^(1/2))`
`u(x)=3x+2 and u'(x)=3`
`v(x)=(2x+1)^(1/2) and v'(x)=1/2*2*(2x+1)^(-1/2)=(2x+1)^(-1/2)`
`f'(x)=(3*(2x+1)^(1/2)-(3x+2)(2x+1)^(-1/2))/(2x+1)`
`f'(x)=(3*(2x+1)-(3x+2))/(2x+1)^(3/2)`
`f'(x)=(3x+1)/(2x+1)^(3/2)`