Question

differentiate w.r.t x.

1+ 3sec^2 x

the answer should be 3sec^2 y tan y

Answer 1

you're wondering how to do it?

Answer 2

yes

Answer 3

do you know the quotient rule?

Answer 4

yes

Answer 5

use it

Answer 6

I assume you're supposed to show how you do it instead of just magically coming up with the answer

Answer 7

so there is no substitution involved

Answer 8

nope

Answer 9

what will be my denominator

Answer 10

isnt there a rule for sec(x), and a power rule for ^n and a chain rule as well?

Answer 11

you can do it using 3/sin^2

Answer 12

it should be 3/cos^2 x

Answer 13

sec = 1/cos ... not seeing the 1/sin^2

Answer 14

sorry, my mistake

Answer 15

ok give me 2 minutes

Answer 16

we learn by mistakes ... and ive made plenty :)

Answer 17

lol am making one now lol

Answer 18

the quotient rule will work ... i just think its more hassle than its worth for this problem.

Answer 19

numerator i have -6cosx

Answer 20

denominator cos^4 x

Answer 21

i have made a slight mistake in the question i don't know if it significant the equation is 1+3 sec^2 y

Answer 22

so implicit differentiation

Answer 23

implicit is just the same stuff ... but a chain rule pops out a y'

Answer 24

the 1 is useless since its a constant; and constants derive to 0

3 sec^2 (y) might be seen better as: (sec y) ^2

run a power rule: 3*2 (sec y), and out pops the insides

6 (sec y) * (sec y)'

the derivatie of sec is sectan

6 sec y * sec y tan y and out pops the insides ..

6 sec y * sec y tan y * y'

Answer 25

the answer is given as 3 sec^2 y tan y

Answer 26

im not all that concerned with "what the answer given" is .... the rules of derivatives dont change regardless of "what the given answer" is.

Answer 27

make sure youve posted the correct information, thats all i have to go off of

Answer 28

ok let me post the whole question

Answer 29

if sin y= 2 sin x
show that (a) (dy/dx)^2= 1+ 3sec^2 x
(i have solved this part as you need to differentiate and then substitute

by differentiating (a) w.r.t.x show that d^y/dx^2=3sec2^y tan y

Answer 30

error on a should be (dy/dx)^2=1+ 3 sec^2 y

Answer 31

sin y= 2 sin x

cos(y) y' = 2 cos(x)

y' = 2 cos(x)/cos(y)

y' = 2 cos(arcsin (sin(y)/2) )/cos(y)

that looks fun

Answer 32

how did you go about that?

Answer 33

firstly we do the substitution for sin y=sqrt 1-cos^2 y

Answer 34

likewise for 2 sin x=sqrt 1-cos^ 2 x

Answer 35

at the end you should get the expression 4 cos^2 x= 3+cos^2 x

Answer 36

\[sin(y)=u\]
\[cos(y)~y'=u'\]
\[y'=\frac{u'}{cos(y)}\]
\[y'=\frac{u'}{\sqrt{1-(u)^2}}\]
\[y'=\frac{2~cos(x)}{\sqrt{1-(2~sin(x))^2}}\]
thats ringing a few bells :)

Answer 37

you the differentiate the original function implicitly and you end up with dy/dx=4cos^x/cos^y then substitute

Answer 38

finally you should end up with 3/cos^2 y + 1 which gives you your answer to part a

Answer 39

that looks fine to me :)

and then theres this part: d^y/dx^2 = 3sec2^y tan y , this is odd notation

Answer 40

sorry am a bit exhausted d^2 y/dx^2=3sec^2 y tan y

Answer 41

so the second derivative; and do they want you to use 1+3sec^2 y for it? or the results of the first derivativing?

Answer 42

yes you are just asked to find the second derivative of the function

Answer 43

\[y'=\frac{2~cos(x)}{\sqrt{1-(2~sin(x))^2}}\]
\[y'=2~cos(x)(1-4~sin^2(x))^{(-1/2)}\]
\[y''=-2~sin(x)(1-4~sin^2(x))^{(-1/2)}+8cos^2(x)sin(x)~(1-4~sin^2(x))^{(-3/2)}\]
looks to be accurate

Answer 44

\[y''=\frac{-2~sin(x)\sqrt{1-4~sin^2(x)}+8cos^2(x)sin(x)}{(1-4~sin^2(x))~\sqrt{1-
4~sin^2(x)}}\]
lol, thats looks to be fun to process

Answer 45

http://www.wolframalpha.com/input/?i=2nd+derivative+arcsin%282sin%28x%29%29

might help to see it simplified

Answer 46

thanks