How do you find the derivative of #y= ln ((x^2 (x+1))/ (x+2)^3)#?

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Answer 1 · 2017-01-09T14:39:11

# dy/dx=(5x+4)/{x(x+1)(x+2)}#

#y=ln{(x^2(x+1))/(x+2)^3}#

Using the Rules of #Log# function, we have,

#y=lnx^2+ln(x+1)-ln(x+2)^3#

#=2lnx+ln(x+1)-3ln(x+2)#

Diff.ing the L.H.S., we will use the Chain Rule.

E.g., #d/dx{ln(x+1)}=1/(x+1)d/dx(x+1)=1/(x+1)#

#:. dy/dx=2(1/x)+1/(x+1)-3/(x+2)#

#={2(x+1)(x+2)+x(x+2)-3x(x+1)}/{x(x+1)(x+2)}#

#:. dy/dx=(5x+4)/{x(x+1)(x+2)}#