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

1 Answer
Jun 16, 2017

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

Explanation:

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)