Question

f(x)+x^4[f(x)]^3=1028 and f(2)=4, find f'(2)

Answer 1

differentiate implicitly

Answer 2

f'(x) + 4x^3*(f(x))^3 + 3(f(x))^2*f '(x)+x^4 = 0

Answer 3

Plug in x = 2:
f '(2) + 4*8 * 4^3 + 3 * 4^2 * f'(2) + 2^4 = 0

Answer 4

f'(2) + 2048 + 48 f'(2) + 16 = 0

Answer 5

49f'(2) = -2048-16
49 f'(2) = -2064
f'(2) = -42.2.. hmm i might've made a careless mistake somewhere, jsut check

Answer 6

thanks very much for the help bahrom, can you help me with this last one
use implicit differentiation to find equation of tangent line to -1x^2+3xy+4y^3=-8 at the point (4,2)

Answer 7

in y=mx + b form

Answer 8

i wonder what ash has been typing lol... let's let him finish

Answer 9

he left wow

Answer 10

my software rejected the -42.2 so i guess i would have to check

Answer 11

can you help me with the other question i posted pls

Answer 12

yea sure.. for the first one i might have made some careless mistake, just check the product rule and multiplication and values

Answer 13

We have
\[f(x)+x^4[f(x)]^3=1028\]
Let's differentiate above equation with respect to x
\[f'(x)+4x^3{f(x)}^3+3x^4 f(x)^2 f'(x)=0\]
let's substitute x=2
we get
\[f'(2)+4\times 2^3 f(2)^3 +3*2^4f(2)^2f'(2)=0\]
f(2)=4
so we get
\[f'(2)+4\times8 \times 64+3\times 16\times 16 f'(2)=0\]
we have
\[f'(2)+2048+4096 f'(2)=0\]
\[4097 f'(2)=-2048\]
\[f'(2)=-2048/4097\]

Answer 14

3(f(x))^2*f '(x)*x^4 it was supposed to be times..

Answer 15

-1x^2+3xy+4y^3=-8 at the point (4,2)
-2x + 3xy'+3y + 12y^2 y' = 0

Answer 16

3xy' + 12y^2y' = 2x - 3y
y' (3x+12y^2) = 2x - 3y
y' = (2x-3y)/(3x+12y^2)
y' = (8 - 6)/(12+48) = 2/60 = 1/30

Answer 17

y = (1/30)x + b

Answer 18

now we just gotta figure out when what y is when x=0

Answer 19

for that use:
(y-y1)/(x-x1) = 1/30
(4-y1)/(2-0) = 1/30
4-y1 = 1/15
4 - 1/15 = y1
y1 = 59/15
y = (1/30)x+(59/15)
Again i hate the numbers, check for careless mistakes

Answer 20

neither work, but i will go back to check

Answer 21

ughhh hold on.. ill write this out.. this is annoying

Answer 22

ok

Answer 23

For the first one try -2048/769

Answer 24

yea that works, thanks

Answer 25

okay.. lol i thought i was going mad..

Answer 26

lol

Answer 27

all right the slope of the 2nd one is correct

Answer 28

ok. my software is rejecting. perhaps, its just how am I typing it in. I may need to figure out how to correctly input it.

Answer 29

is it called webworks? lol

Answer 30

no i might be doing something wrong

Answer 31

yea, does your school use it as well?

Answer 32

well i was lucky enough to skip it.. but i used to solve my friends' webworks for them lol

Answer 33

lucky you. i hate this thing. sometimes correctly entering your solution is more difficult than solving the actual problem

Answer 34

i know.. i once decided to round 2.999 up to 3... god the answer it finally accepted was 2.999

Answer 35

lol i have encounter that issue alot of times

Answer 36

found my mistake

Answer 37

nice. where did you go wrong?

Answer 38

4-y1 = 1/15
4 - 1/15 = y1
y1 = 59/15
y = (1/30)x+(59/15)

i used x instead of y and y instead of x.

Answer 39

oh ok

Answer 40

y = (1/30)x+(28/15) <- your answer

Answer 41

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