Let #y=[(x^(2x)(x-1)^3)/(3+5x)^4]#, how do you use logarithmic differentiation to find #dy/dx#?

1 Answer
Jan 12, 2017

# dy/dx= (x^(2x)(x-1)^3)/(3+5x)^4{2 + 2lnx + 3/(x-1)-20/(3+5x)} #

Explanation:

#y=[(x^(2x)(x-1)^3)/(3+5x)^4]#

Taking (natural) logs of both sides:

# ln y=ln[(x^(2x)(x-1)^3)/(3+5x)^4] #
# \ \ \ \ \ \=ln \ x^(2x) + ln\ (x-1)^3 - ln \ (3+5x)^4 #
# \ \ \ \ \ \=2xln \ x + 3ln\ (x-1) - 4ln \ (3+5x) #

Differentiating the LHS implicitly and the RHS using the product rule gives:
# 1/y dy/dx \ = (2x)(1/x) + (2)(lnx) + 3/(x-1)-4*5/(3+5x) #
# \ \ \ \ \ \ \ \ \ \ = 2 + 2lnx + 3/(x-1)-20/(3+5x) #
# :. dy/dx= (x^(2x)(x-1)^3)/(3+5x)^4{2 + 2lnx + 3/(x-1)-20/(3+5x)} #