What is the derivative of #ln[(2x-3)/(7x+8)]^(1/2)#?

1 Answer
May 2, 2018

The answer #(1/(2x-3))-(7/(14x+16))#

Explanation:

show below we will use the properties of ln

#ln[(2x-3)/(7x+8)]^(1/2)=1/2*ln[(2x-3)/(7x+8)]#

#1/2*ln[(2x-3)/(7x+8)]=1/2[ln(2x-3)-ln(7x+8)]#

#1/2ln(2x-3)-1/2ln(7x+8)#

now the derivative

#1/2*(2/(2x-3))-1/2*(7/(7x+8))#

#(1/(2x-3))-(7/(14x+16))#