Question

y=2sinx, then what would be the derivative for this?

Answer 1

whats derivative of sin x ?

Answer 2

cosx

Answer 3

2 can be pulled out of differentiation,
d/dx (2 sin x) = 2 d/dx (sin x) = 2 cos x
thats it!

Answer 4

so we just leave the 2 alone...? so then what are you allowed to pull out of a differentiation? definitely not a variable

Answer 5

a constant
2 is a constant

Answer 6

\((d/dx) a f(x)= a (d/dx)f(x)=af'(x)\)

Answer 7

so technically you could factor a constant out of a function being differentiated then take the derivative of the factored function, then multiply the constant back in?

Answer 8

thats correct.

Answer 9

but if it was something like 2+sin x
then derivative of constant = 0
so, that would be 0+cos x = cos x

Answer 10

oh I c ok because you have to take the derivatives seperately~ differentiation and derivative is the same thing?

Answer 11

yes, its same thing.

Answer 12

ok so what aboutfinding the derivative of sinx^2???

Answer 13

\(\sin (x^2)\)
right ? chain rule...

Answer 14

OH ic~ ok thanx!

Answer 15

what u got for sin (x^2)
just to verify

Answer 16

2xcosx^2?

Answer 17

correct :)

Answer 18

oh, and how to deal with secant?

Answer 19

d/dx (sec x) = sec x tan x

Answer 20

how...?

Answer 21

you want me to prove that ?
write sec x = 1/cos x
d/dx (1/cos x) =... ?
use chain rule

Answer 22

chain??? you mean 1(-sinx)+cosx(0)... wait that doesn't work...? hmm how to use chain rule for this...? (cosx)^-1, -cosx?????

Answer 23

whats derivative of 1/x ?

Answer 24

1(1)+x(o)
1?

Answer 25

was that product rule you just applied ? :P

Answer 26

derivative of x^{-1} is .... ?
use the formula for derivative of x^n

Answer 27

oops LOL uh, [x(0) + 1(1)]/x^2
1/x^2

Answer 28

oh, so you tried quotient rule now....there's - *minus* in middle

Answer 29

i would have used that d/dx (x^n) = n x^{n-1}
d/dx (x^{-1}) = -1 x^{-1-1} = -1/x^2

Answer 30

omg I must be tired

Answer 31

oh, ok, Ic

Answer 32

so, d/dx (1/cos x) = -1/ (cos^2 x) * (d/dx (cos x)) [using chain rule]
= -1/ (cos^2 x) * (-sin x)
= sin x / cos^2 x
= (sin x/ cos x) * (1/ cos x)
= tan x * sec x
ask if any doubts....

Answer 33

how did u get d/dx(cosx) in that first step?
and the original question was like...d^2/d^2x just so u kno I dont understand what that notation means etiher...LOL

Answer 34

d/dx (cos x) means derivative of cos x which is sin x

Answer 35

d^2/d^2x = d/dx (d/dx )
means double derivative.
taking derivative twice, one after another

Answer 36

like d^2/d^2x (x^3) = d/dx (d/dx x^3) = d/dx (3x^2 ) = 6x

Answer 37

oh ok so then I understand how you got the first derivative now. You saw it as (cos)^-1, which allowed for the chain rule ic. so then after tanx times secx, now it is the problem of geeting the second derivative.

Answer 38

means you had to find d^2/d^2x (sec x) ??

Answer 39

si, that would be correct. So then I tried it, and I got -secx

Answer 40

so, now you need d/dx (sec x tan x)
used product rule ?
(how come the answer is so compact ? i don't think thats correct...)

Answer 41

my answer is probs wrong... is not one of the answer choices on here...

Answer 42

(sinx/cosx)(sinx/cosx times 1/cosx) + {1/cosx[(cosx times cosx)-(sinx times (-sinx))]}/cos^2x

sin^2x/cos^3x+(cos^2x+sin^2x)/cos^3x)

sin^2x+1/cos^3x

-cos^2x/cos^3x

-1/cosx

-secx

Answer 43

upto here its correct
sin^2x+1/cos^3x

Answer 44

how 1+sin^2 x = -cos^2 x ?

Answer 45

oh!!! I think I was thinking it was sin^2x -1 LOL

so then where do I go from there?

Answer 46

you can leave it like this :(sin^2x+1)/cos^3x
if there's such an option...
or ((sin^2 x/ cos^2x) * (1/ cos x))+ 1/ cos^3 x = tan^2 x sec x + sec^3 x

Answer 47

if you still didn't get the required answer, list the choices.

Answer 48

got it~! oh, and I have another question, but I'm goihng to open up another question cya there?

Answer 49

sure :)