Question

find derivative of (secx)/(1+cosx) please show steps.

Answer 1

\[\frac{(1+cosx)\times (secx)' - secx ( 1 +cosx)'}{(1 + cosx)^2}\]

Answer 2

(secx)' = secx * tanx
(1 + cosx)' = -sinx

Answer 3

yeah im still stuck after that. I tried factoring out sec

Answer 4

secx*tanx(1 +cosx) + secx*sinx
---------------------------
(1 + cosx)^2

secx(tanx + tanx*cosx + tanx)
--------------------------
(1 + cosx)^2

secx(2tanx + sinx)
------------------
(1 + cosx)^2
2tanx*secx + tanx
---------------
(1 +cosx)^2


I guess this much simplification must be enough

Answer 5

uhm on the second step what happened to sinx?

Answer 6

sinx*secx = sinx*1/cosx = sinx/cosx =tanx

Answer 7

ishaan didn't you factor the sec(x) out though?

Answer 8

yeah thats what i was thinking to myinanaya

Answer 9

\[(\frac{\sec(x)}{1+\cos(x)})'=\frac{(\sec(x))'(1+\cos(x))-\sec(x)(1+\cos(x))')}{(1+\cos(x))^2}\]
\[=\frac{\sec(x)\tan(x)(1+\cos(x))-\sec(x)(0-\sin(x))}{(1+\cos(x))^2}\]
\[=\frac{\sec(x)\tan(x)+\sec(x)\tan(x)\cos(x)+\sec(x)\sin(x)}{(1+\cos(x))^2}\]
\[=\frac{\sec(x)\tan(x)+\tan(x)+\tan(x)}{(1+\cos(x))^2}\]
\[=\frac{\sec(x)\tan(x)+2\tan(x)}{(1+\cos(x))^2}\]
\[=\frac{\tan(x)(\sec(x)+2)}{(1+\cos(x))^2}\]

Answer 10

on your thrid step, what happened to secxtanxcosx? cause a tan is replaced there

Answer 11

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

Answer 12

\[\sec(x)=\frac{1}{\cos(x)}\] right?

Answer 13

so \[\frac{1}{\cos(x)} \cdot \cos(x)=?\]

Answer 14

1

Answer 15

so what is 1*tan(x)?

Answer 16

tanx -___- okay I see. THANKS U SO MUCH!. do u think u can help me on one more please?

Answer 17

i have to go to bed
its late
and i have to wake up at 5 am :(

Answer 18

awesome latex

Answer 19

ishaan can help right?

Answer 20

I will try
goodnight myininaya

Answer 21

you will do fine
but you did make one error above did you see it?

Answer 22

good night myininaya!

Answer 23

goodnight and good luck you guys! I believe in you!

Answer 24

error hmm no I didn't what is it?

Answer 25

well i see one error and i stop reading after that

Answer 26

ok i'm going for real

Answer 27

okay so what is the derivative of (8x-4)(2x^-2+x^-4)

Answer 28

(8x-4)'*(2x^(-2) + x^(-4)) + (2x^(-2) + x^(-4))'*8x-4
you can do it

Answer 29

I actually tried for 2x^-2+x^-4 I got for the derivative -4/x^3 -4/x^5...is that correct?

Answer 30

oh yeah I see it now @myininaya thanks

Answer 31

it is correct, good job

Answer 32

okay so I have (8x-4)((5x^7+1)+(2x^-2+x^-4)(8) correct?

Answer 33

correct

Answer 34

then i have: (-32/x^3+16x^5)+(16x^-2+8x^-4)

Answer 35

except (8x -4)(5x^7 +1) part how'd you get this

Answer 36

oh my bad it was suppose to be times -4/x^3 -4/x^5

Answer 37

you are right then

Answer 38

(-32/x^3+16x^5)+(16x^-2+8x^-4) correct too right?

Answer 39

(8x -4)(-4/x^3 -4/x^5) = -32/x^2 - 32/x^4 + 16/x^3 + 16/x^5



-32/x^2 - 32/x^4 + 16/x^3 + 16/x^5 + (16/x^2+8/x^4)

Answer 40

reply to
'(-32/x^3+16x^5)+(16x^-2+8x^-4) correct too right?'




check it again doesn't seem correct to me

Answer 41

yeah i forgot to foil and put the x's on the denominator. okay so after that, what happens to the x denominators?

Answer 42

-32/x^2 - 32/x^4 + 16/x^3 + 16/x^5 + (16/x^2+8/x^4)
-16/x^2 - 24/x^4 + 16/x^3 + 16/x^5

Answer 43

uhm how is it -24/x^4?

Answer 44

oh wiat nvm

Answer 45

thx

Answer 46

no problem