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

Question

Christopher Allers · Accepted Answer

# (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) #

Emma Aldridge · Answer

undefined To find the derivative of $ z = \ln(y\sqrt{3y + 1}) $ with respect to $ y $, you would use the chain rule. The derivative is $ \frac{dz}{dy} = \frac{1}{y\sqrt{3y + 1}} \left( \frac{d}{dy}(y\sqrt{3y + 1}) \right) $. Then, differentiate $ y\sqrt{3y + 1} $ using the product rule, which results in $ \frac{dz}{dy} = \frac{1}{y\sqrt{3y + 1}} \left( \sqrt{3y + 1} + \frac{y}{2\sqrt{3y + 1}} \cdot 3 \right) $. Simplify this expression to get the final derivative.