How do you find the derivatives of #z=ln(ysqrt(3y+1))#?

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Answer 1 · 2017-06-16T12:20:18

# (dz)/(dy) = {9y+2}/(6y^2+2y) #

We have:

# z = ln( ysqrt(3y+1)) #

Using the rules of logs:

# log ab = loga+logb #, and, #log a^b=bloga#

we can rewrite the expression as:

# z = lny + ln sqrt(3y+1) #
# \ \ = lny + ln (3y+1)^(1/2) #
# \ \ = lny + 1/2 \ ln (3y+1) #

Then using the standard Calculus result:

# d/dx lnx = 1/x #

along with the chain rule, we have:

# (dz)/(dy) = 1/y + 1/2 * 1/(3y+1) * 3#
# " " = 1/y + 3/2 * 1/(3y+1) #

We could if required simplify this further:

# (dz)/(dy) = {2(3y+1) + 3y}/(2y(3y+1)) #
# " " = {6y+2 + 3y}/(6y^2+2y) #
# " " = {9y+2}/(6y^2+2y) #