How do I find the derivative of # ln[x(x^2+1)^2/(2x^3-1)^(1/2)] #?
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"How do you find the derivative of #y=ln(5x)#?"
Use the properties of logarithms and the #d/dxln(x)# rule to get #1/x+(4x)/(x^2+1)-(3x^2)/(2x^3-1)#.
Begin by using the properties of logs to write:
#ln[(x(x^2+1)^2)/(2x^3-1)^(1/2)]=ln(x(x^2+1)^2)-ln(2x^3-1)^(1/2)=ln(x)+ln((x^2+1)^2)-1/2ln(2x^3-1)=ln(x)+2ln(x^2+1)-1/2ln(2x^3-1)#
Now we can take the derivative of this term by term.
Term 1: #ln(x)#
This is the easiest, as the derivative of #ln(x)=1/x#.
Term 2: #2ln(x^2+1)#
We need to use the chain rule for this one. We first apply the derivative of #ln(x)# rule to get #2*1/(x^2+1)#, and then we apply the chain rule to get #2*(2x)/(x^2+1)=(4x)/(x^2+1)#.
Term 3: #1/2ln(2x^3-1)#
We use the same process as term 2. Use #ln(x)# rule to get #1/2*1/(2x^3-1)#, then use the chain rule to get #1/2*(6x^2)/(2x^3-1)=(3x^2)/(2x^3-1)#.
Put all of these together to get the final result:
#d/dxln[(x(x^2+1)^2)/(2x^3-1)^(1/2)]=1/x+(4x)/(x^2+1)-(3x^2)/(2x^3-1)#.
