How do you differentiate #ln(6z^4+3z^2)#, #ln(6x^2+3z)#, and #ln(4z+1)# assuming #z0#?

Question

How do you differentiate #ln(6z^4+3z^2)#, #ln(6x^2+3z)#, and #ln(4z+1)# assuming #z>0#?

Paisley Johnson · Accepted Answer

#(24z^3+6z)/(6z^4+3z^2)#, #(12z+3)/(6z^2+3z)# (I assume you didn't mean to put #x# there?) and #4/(4z+1)# or #1/z+1# nb: For all of these I'm assuming they are #y=ln("function")# and we are differentiating y with respect to z: #dy/dz# Rule for natural log is #color(red)"derivative"# on top, #color(blue)"function"# on the bottom. 1) #ln(6z^4+3z^2)# you get #color(blue)(6z^4+3z^2# on the bottom, and its derivative, #color(red)(24z^3+6z#, on top. This is just a normal polynomial derivative found by multiplying the coefficient by the power, then taking 1 off the power. #color(red)(24z^3+6z)/color(blue)(6z^4+3z^2)# 2) #ln(6z^2+3z)# if you did mean #6x^2# instead of #6z^2#, I'm not sure how to solve it unless you know whether these are #y=# or #x=# functions. This is still a #ln# function so the same rule applies. #color(blue)"Function"# on the bottom: #color(blue)(6z^2+3z# and its #color(red)"derivative"#: #color(red)(12z+3# on top. #color(red)(12z+3)/color(blue)(6z^2+3z)# 3) #ln(4z+1)# #color(red)"Derivative"# of what's in the bracket is just #color(red)4# because dividing out a #z# from #4z# just gives 4, and the 1 is of no power so goes. #:. dy/dz ln(4z+1)# = #color(red)4/color(blue)(4z+1)# this simplifies down: #cancel(4)/(cancel(4)z+1)#=#1/(z+1) = 1/z +1#

How do you differentiate #ln(6z^4+3z^2)#, #ln(6x^2+3z)#, and #ln(4z+1)# assuming #z&gt;0#?

How do you differentiate #ln(6z^4+3z^2)#, #ln(6x^2+3z)#, and #ln(4z+1)# assuming #z>0#?