Question

find the derivative
y = ln(8x^3 + 3^8x)

Answer 1

its absolute value

Answer 2

ln absolute value

Answer 3

\[y=\ln \left| 8x ^{3}+3^{8x} \right|\]

Answer 4

how would you input this on wolphram.com

Answer 5

unless one of you can solve this

Answer 6

like this: ln|8x^3 + 3^8x|

Answer 7

derive ln|8x^3 + 3^8x| for the derivative.

Answer 8

unless one of you can solve this here

Answer 9

note:

if \[f(x)>0\]

\[\frac{d}{dx}\ln(|f(x)|)=\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}\]

if

\[f(x)<0\]

\[\frac{d}{dx}\ln(|f(x)|)=\frac{d}{dx}\ln(-f(x))=\frac{-f'(x)}{-f(x)}=\frac{f'(x)}{f(x)}\]

Answer 10

forgot peranteces but its good thomas

Answer 11

\[y=3^{8x}\]
\[lny=\ln(3^{8x})\]
\[lny=8xln3\]
\[lny=8(\ln3)x\]
\[\frac{y'}{y}=8\ln3\]
\[y'=y*8\ln3\]
\[y'=3^{8x}*8\ln3\]

Answer 12

\[\frac{dy}{dx} = \frac {1}{8x^3 +3^{8x}}\times ( 24x^2 + \frac{d}{dx} 3^{8x} )\]

Answer 13

then for the part second

\[f = 3^{8x}\]\[\ln f = 8x \ln 3\]\[\frac{1}{f} \frac{df}{dx} = \ln3 \times 8\]then what you get \[\frac{df}{dx} = 3^{8x} \times \ln3\times 8\]Now all you need to do is to replace the Values .. : )

Answer 14

\[\frac{dy}{dx} = \frac{1}{8x^2 + 3^{8x} }\times (24x^2 + 3^{8x}\times \ln3 \times 8)\]should be the derivative ... i'm doing it so long because I learned this fraction today and I like it ...lol :D