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

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Answer 1 · 2018-05-02T01:05:35

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

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))#