plz...help me with this mathematics problems.?I really need your help?

Question

first question)

find dy/dx of f(x) in the folowing function

f(x)=(x+4)^2(2-5x)^3


second question;)
Log base 8 (1/32)+3 Log base 8 (4) divided by Log base 8 (64)=

plz...I really appreciate that...
well,happy new year........

Answer 1

y= f(x) = (x+4)^2(2-5x)^3
log y = 2log(x+4) + 3log(2-5x)

(1/y)dy/dx = (2/(x+4)) - (3/(2-5x))

dy/dx = [(x+4)^2(2-5x)^3] / {[2/(x+4)) - (3/(2-5x)]}
____________ __________ ___________ ________

Log base 8 (1/32)+3 Log base 8 (4) divided by Log base 8 (64)
=[(-log 32)/log 8 + 3*log 4/log8 ] {log 64/log8}
= (- log 32 + log 4^3)/ log 64
= log(64/32) / log 64
= log2/log2^6
=1/6

Answer 2

For the first you need to use product rule and chain rule:

f'(x) = (x + 4)²[(2 - 5x)³]' + [(x + 4)²]'(2 - 5x)³
f'(x) = (x + 4)²(3)(2 - 5x)²(-5) + 2(x + 4)(2 - 5x)³

I'll let you have fun simplifying that. :)

For the second, remember the log is the power to which you raise the base in order to get the number.

1/32 = 1/2^5 = 2^(-5) = (8^1/3)^(-5), so log8 1/32 = -5/3.
8^(2) = 64, so log8 64 = 2.
4 = 2^2 = (8^1/3)^2 = 8^(2/3), so log8 4 = 2/3.

log8 1/32 + 3 log8 4 / log8 64
= -5/3 + 3 (2/3) / 2 = -5/3 + 2/2 = -5/3 + 1 = -2/3

(Unless you missed some parentheses, if you meant:

(log8 1/32 + 3 log8 4)/ log8 64

Then that would be (-5/3 + 2)/2 = (1/3)/2 = 1/6.)

Answer 3

1) Take ln to simplify your work as showed above.

2) None of above is right. Therefore, I only do this one.
(log₈1/32 + 3log₈4)/log₈64
= (log₈64/32)/2
= (log₈2)/2
=1/6

------------------
claYlia a,

They corrected their answers after my post.

Answer 4

y'=2(x+4).1.(2-5x)^3+ 3.(2-5x)^2. (-5).(x+4)^2

log 2^-5 log2^2 log2^3
------------+-------------------. -----------
log2^3 log2^3 log 2^4

=-5/3+2/3. 3/4
=11/6

Answer 5

for the first one use the foil method- (x+4)(x+4) and for the second use distributive i think- 3x2, 3x5x (6-15x) and then go from there. IDK the second one

Answer 6

i have no clue tht seems sooooo hard