Question

find dy/dx of sin(x+y) + sin (x-y) = 1
a. -cotxcoty
b. cotxcoty
c. 1
d. -1

Answer 1

\[(x+y)_xcos(x+y)+(x-y)_x \cos(x-y)=1_x\]
\[(x_x+y_x)\cos(x+y)+(x_x-y_x)\cos(x-y)=0\]
\[(1+y')\cos(x+y)+(1-y')\cos(x-y)=0\]
Distribute
\[\cos(x+y)+y' \cos(x+y)+\cos(x-y)-y'\cos(x-y)=0\]
\[y'(\cos(x+y)-\cos(x-y))=-\cos(x-y)-\cos(x+y)\]
do you think you can finish this?

Answer 2

-1?

Answer 3

\[\frac{-(\cos(x-y)+\cos(x+y))}{\cos(x+y)-\cos(x-y)}\]

Answer 4

so you have to do some trig identity stuff

Answer 5

(-cos(x-y)-cos(x+y))/(y-cos(x)+cos(x+y)) then after this?

Answer 6

\[=-\frac{\cos(x) \cos(y)+\sin(x) \sin(y)+\cos(x) \cos(y)- \sin(x) \sin(y)}{\cos(x) \cos(y) - \sin(x) \sin(y)-\cos(x) \cos(y) - \sin(x) \sin(y)}\]

Answer 7

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

Answer 8

\[=\frac{\cos(x)}{\sin(x)} \cdot \frac{\cos(y)}{\sin(y)}\]

Answer 9

got it from here now?

Answer 10

thats y u said i need trigo identity ic ic from my trigo identity on my note book it is cotxcoty

Answer 11

lol yep
:)

Answer 12

hey phi can yo help
i must leave
sorry guys

Answer 13

thank myininaya

Answer 14

y = (sin2x)/4 + x/2

find dy/dx

Answer 15

x = pi /2 ryt?

Answer 16

that's what I mean
find dy/dx for y= 1/4 sin(2x) +x/2

dy/dx = (1/4) * cos(2x) * 2 *dx/dx + 1/2 dx/dx
= cos(2x)/2 + 1/2

now solve the equation
cos(2x)/2 + 1/2 = 1
or cos(2x)+1 = 2
cos(2x) = 1
2x = acos(1) = 0 or 2 pi
x= 0/2 = 0 or 2 pi/2 = pi
so x= n*pi for n= an integer

Answer 17

the only one that matches is (b) 0

Answer 18

ow 6 people i thought like this also cos(2x)=-1
2x=cos−1(−1)

2x=π

x=π/2

Answer 19

i go for 0 now =)

Answer 20

when in doubt, use wolfram
http://www.wolframalpha.com/input/?i=d%2Fdx+%28sin2x%29%2F4+%2B+x%2F2+%3D+1

Answer 21

thanks again phi