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

2 Answers
Jun 17, 2018

#(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))#

Jun 17, 2018

#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)#