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

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

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Answer 1 · 2017-09-19T20:14:14

$\frac{24 z^{3} + 6 z}{6 z^{4} + 3 z^{2}}$ $(24z^3+6z)/(6z^4+3z^2)$ , $\frac{12 z + 3}{6 z^{2} + 3 z}$ $(12z+3)/(6z^2+3z)$ (I assume you didn't mean to put $x$ $x$ there?) and $\frac{4}{4 z + 1}$ $4/(4z+1)$ or $\frac{1}{z} + 1$ $1/z+1$

Explanation:

nb: For all of these I'm assuming they are $y = ln (function)$ $y=ln("function")$ and we are differentiating y with respect to z: $\frac{d y}{d z}$ $dy/dz$

Rule for natural log is $derivative$ $color(red)"derivative"$ on top, $function$ $color(blue)"function"$ on the bottom.

1) $ln (6 z^{4} + 3 z^{2})$ $ln(6z^4+3z^2)$
you get $6 z^{4} + 3 z^{2}$ $color(blue)(6z^4+3z^2$ on the bottom, and its derivative, $24 z^{3} + 6 z$ $color(red)(24z^3+6z$ , on top. This is just a normal polynomial derivative found by multiplying the coefficient by the power, then taking 1 off the power.
$\frac{24 z^{3} + 6 z}{6 z^{4} + 3 z^{2}}$ $color(red)(24z^3+6z)/color(blue)(6z^4+3z^2)$

2) $ln (6 z^{2} + 3 z)$ $ln(6z^2+3z)$ if you did mean $6 x^{2}$ $6x^2$ instead of $6 z^{2}$ $6z^2$ , I'm not sure how to solve it unless you know whether these are $y =$ $y=$ or $x =$ $x=$ functions.

This is still a $ln$ $ln$ function so the same rule applies. $Function$ $color(blue)"Function"$ on the bottom: $6 z^{2} + 3 z$ $color(blue)(6z^2+3z$ and its $derivative$ $color(red)"derivative"$ : $12 z + 3$ $color(red)(12z+3$ on top.

$\frac{12 z + 3}{6 z^{2} + 3 z}$ $color(red)(12z+3)/color(blue)(6z^2+3z)$

3) $ln (4 z + 1)$ $ln(4z+1)$
$Derivative$ $color(red)"Derivative"$ of what's in the bracket is just $4$ $color(red)4$ because dividing out a $z$ $z$ from $4 z$ $4z$ 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: $cancel(4)/(cancel(4)z+1)$ = $1/(z+1) = 1/z +1$

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

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How do you differentiate ln(6z^4+3z^2)ln(6z4+3z2)ln(6z^4+3z^2), ln(6x^2+3z)ln(6x2+3z)ln(6x^2+3z), and ln(4z+1)ln(4z+1)ln(4z+1) assuming z>0z>0z>0?

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