help w/ differentiate this function: ln(z^8)+ln(8z)+ln(8)+8lnz-ln(e^z)?

Question

The answer comes to: 17/z-1
Thanks a bunch

Answer 1

f(z) = ln(z^8) + ln(8z) + ln(8) + 8lnz - ln(e^z)

f(z) = 8ln(z) + ln(8) + ln(z) + ln(8) + 8ln(z) - z

f(z) = 17ln(z) + 2ln(8) - z

f'(z) = 17/z - 1

Answer 2

the derivative of anything ln(x) is 1/x

So:
1/(z^8) * (8z^7) + 1/(8z) * 8 + 0 + 8/z - 1

Because it's the derivative of the ln(az) times the derivative of the inside piece... And because ln(8) is just a number, so its derivative is 0. Also, ln(e) cancels itself out, just as e^ln cancels out, so the last part is simply z, and the derivative of z is 1.

Now when you simplify it, you get
8/z + 1/z + 8/z - 1

= 17/z - 1

Answer 3

Let's do it term by term, using the chain rule
and properties of logs
d/dz ln(z^8) = d/dz 8ln(z) = 8/z.
d/dz ln(8z) = d/dz (ln 8 + ln z) = 1/z
d/dz ln(8) = 0 because ln(8) is a constant.
d/dz(8lnz) = 8/z
-d/dz ln(e^z) = -d/dz (z) = -1
Putting it all together, the answer is 17/z -1.

Answer 4

Use properties of ln:

ln(a^b) = b*ln(a)
ln(a*b) = ln(a) + ln(b) (if a and b are both >0)
ln(e^x) = x

ln(z^8)+ln(8z)+ln(8)+8lnz-ln(e^z)
= 8*ln(z) + ln(8) + ln(z) + ln(8) + 8ln(z) - z
= 17ln(z) + 2*ln(8) - z

d ln(z)/dz = 1/z,
d z/dz = 1
d constant/dz = 0
=> d [ln(z^8)+ln(8z)+ln(8)+8lnz-ln(e^z)]/dz =
d[17lnz - z+ 2*ln(8) ]/dz = 17/z -1

Answer 5

y= ln(z^8)+ln(8z)+ln(8)+8lnz-ln(e^z)
y= 8ln(z)+ln(8z)+ln(8)+8lnz-zln(e)
dy/dz=8(1/z) +(1/8z)8 + 0 + 8(1/z) -1(lne)........log(e)e=1
dy/dz=8/z +1/z +8/z -1(1)
dy/dz=( 8 +1 + 8 )/z -1
dy/dz= 17/z -1
qed (quite easily done)