Question

find the equation to the tangent line to the curve cosy+ycosx = y+sin(x^2)+sin^2(x) at the point (0,pi/2)

Answer 1

start taking the derivatives ....

Answer 2

is this format correct? : y=cosy+ycosx-sin(x^2)-sin^2(x)

Answer 3

dont "solve for y"; just jump in and start deriving the parts

Answer 4

cosy: -y' sin(y)
ycosx: y' cosx - y sinx
=
y: y'
sin(x^2): 2x cos(x^2)
sin^2(x): 2sinx cosx

now plug in the x,y point given and solve for y'

Answer 5

so it would be: y'=-siny-ysinx-2xcos(x^2)-2sinxcosx

Answer 6

looks about right, hard to see thru all the trig, which is why idda just inputed x,y as is to reduce the cultter

Answer 7

-y' sin(y) + y' cosx - y sinx = y' + 2x cos(x^2) + 2sinx cosx

-y' sin(pi/2) + y' cos0 - pi/2 sin0 = y' + 2.0 cos(0) + 2sinpi/2 cos0

-y' 1 + y' 1 - pi/2 0 = y' + 0 + 2.1.1

0 = y' + 2

Answer 8

the slope of your line is equal to y'; so -2, and they give you the point to anchor it top

Answer 9

*anchor it to

Answer 10

I got my slope to be 0.

Answer 11

then one of us messed up :)

Answer 12

because I used my y' equation and plugged 0 into it since that's the x coordinate

Answer 13

and my y=pi/2

Answer 14

\[-y' sin(y) + y' cosx - y~sinx = y' + 2x~cos(x^2) + 2sinx~cosx\]
\[-y' sin(y) + y' cosx -y' = y~sinx + 2x~cos(x^2) + 2sinx~cosx\]
\[y'(-sin(y) + cosx -1) = y~sinx + 2x~cos(x^2) + 2sinx~cosx\]
\[y'=\frac{ y~sinx + 2x~cos(x^2) + 2sinx~cosx}{-sin(y) + cosx -1 }\]

Answer 15

your y' was off id say

Answer 16

\[y'=\frac{ \frac{pi}{2}~sin0 + 2(0)~cos(0) + 2sin0~cos0}{-sin(\frac{pi}{2}) + cos0 -1 }\]
\[y'=\frac{ \frac{pi}{2}~0 + 0 + 2(0)}{-1 +1 -1 }\]
pfft, now i get a zero lol

Answer 17

-y' sin(y) + y' cosx - y sinx = y' + 2x cos(x^2) + 2sinx cosx

0 = y' + 2sin pi/2 cos0 <--- i used y insted of x for that one ...

y' = 0