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

1 Answer
Sep 19, 2017

(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

Explanation:

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