Question

Hey I need help with some calculus questions. (:

Answer 1

ask @Michele_Laino he can prob help

Answer 2

hii

Answer 3

Hey can you help me?

Answer 4

Yes

Answer 5

I already found the derivatives of each function. I just dont understand how to find the domain.

Answer 6

when x<4, f(x)=5x-6 whose derivative is 5
when 4<=x<=6 f(x)=x^2-2 whose derivative is 2x
when x>6, f(x)=4x+10 whose derivative is 4

Answer 7

now we have to check the domain

Answer 8

Yes, I got that part (:

Answer 9

f'(4)=2.4=8 which is not equal to the Lf'(4)

Answer 10

Wait. how did you do that.

Answer 11

f'(6)=2.6=12 and Rf'(6)=4 so f is not derivable at x=6

Answer 12

for existence of derivative we must have Lf'(x)=Rf'(x)

Answer 13

Okay, I am still a little confused on that part, so the domain is that x cannot equal 4 and 6

Answer 14

wait

Answer 15

Well all real numbers but 4 and 6

Answer 16

no no let me write the answer
f'(x)=5, when x<4
f'(x)=2x, when 4<x<6
f'(x)=4 , when x>6

Answer 17

Ohhhhh okay. I see now.

Answer 18

yeah the domain is all real numbers except 4 and 6

Answer 19

Thank you. That makes sense!

Answer 20

Would you mind trying to help me with another question?

Answer 21

ok

Answer 22

Getting stuck on this one

Answer 23

multiply numerator and denominator by 1+cos x

Answer 24

so
\[y=\frac{ (1+\cos x)^2 }{ 1-\cos ^2 x}=\frac{ 1+\cos ^2 x +2 \cos x }{ \sin ^2 x }\]

Answer 25

wait you may directly differentiate

Answer 26

See, the answer choices are confusing me. I cant figure out how to get one of those.

Answer 27

\[\frac{ d }{ dx}(\frac{ f(x) }{ g(x) })=\frac{ g(x)f'(x)-f(x)g'(x) }{ {g(x)}^2 }\]

Answer 28

@arianna1453

Answer 29

oh okey.

Answer 30

use the formula and tell me the answer

Answer 31

@arianna1453 option d is correct

Answer 32

Give me a second. Working it out.

Answer 33

are you sure its d @jango_IN_DTOWN ?

Answer 34

yeah you want me to work it out?

Answer 35

Yeah, if you could.

Answer 36

\[f'(x)=\frac{ (1-\cos x)(-\sin x)-(1+\cos x) \sin x }{ (1-\cos x)^2 }\]

Answer 37

\[=\frac{ -\sin x +\sin xcos x-\sin x -\sin x \cos x }{ (1-\cos x)^2 }\]

Answer 38

\[=\frac{ -2\sin x }{ (1-\cos x)^2 }\]

Answer 39

OH. I had one of my signs wrong. I got it!! Thank you

Answer 40

n.p.

Answer 41

Last one i will ask you. Is this one correct?

Answer 42

@jango_IN_DTOWN ?

Answer 43

option c is correct