Question

What is the slope of the line tangent to the curve y+2 = (x^2/2) - 2siny at the point (2,0)?

Answer 1

y + 2 = (x^(2)/2)
rearange the equation take the derivative

Answer 2

i know the derivative is teh slope of the line to a tangent but m not sure how to differentiate that

Answer 3

u forgot the -2siny

Answer 4

sorry f'(x)(x-a) + f(x) = m

Answer 5

i know the equation but just not how 2 differentiate f(x)....tahts wat i need help with

Answer 6

alright, y+2 = (x^2/2) - 2sin(y)

so we are taking the derivative y in respect to x so we have
dy/dx use chain rule on y

so y' = 2x/2 - 2cos(y)*y'

Answer 7

the 2 disapears cuz the derivative of a constant is 0, right?

Answer 8

Yes to your question

Now rearrange it to solve for y'

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

0 = x - 2cos(y)y' - y'
- x = 2cos(y)y' - y'
-x = y'(2cos(y) - 1)
-x/(2cos(y) - 1) = y'

Answer 9

we know when f(2) = 0 so thus y = 0
so when

f'(2) = -2/(2cos(0)-1)

Answer 10

what happened to the 2x/2 after u made it equal to 0? why did it cancel?

Answer 11

2/2 = 1

Answer 12

x(2/2) = x

Answer 13

OH righ? sorry i see it now

Answer 14

when i plug in 0 i get -2/0 = 0

Answer 15

you cant divide by 0

Answer 16

so what do i do?

Answer 17

f'(2) = -2/(2cos(0)-1)
cos(0) = 1

thus
f'(2) = -2/(2(1)-1)
= -2/-1
= 2

f'(2) = 2

Answer 18

how did u get cos(0) =1

Answer 19

I have the unit circle memorized

Answer 20

cos = x
sin = y

Answer 21

i have the asnwer key 2 the question but its says the asnwer is 2/3...its just taht idk how to get that answer

Answer 22

I made a mistake and didnt carry the negative

f'(2) = -2/(-2cos(0)-1)
f'(2) = 2/3

Answer 23

that is your answer

Answer 24

y' = 2x/2 - 2cos(y)*y'
0 = x - 2cos(y)y' - y'
- x = - 2cos(y)y' - y'
- x = y'(- 2cos(y) - 1)
- x/(- 2cos(y) - 1) = y'

Answer 25

YOU ARE AWESOME! I TOTALLY UNDERSTOOD THE WHOLE PROCESS. Sorry 4 askingso many questions. Sometimes i dont see things right away

Answer 26

its fine engagement is a good thing