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)