Question

Use implicit diffentiation to find an equation of the line tangent to the given curve at the point (2,1)

sin(xy-2)+x-2=(y-1)cos(x-2y)

@perl :)

Answer 1

Differentiation*

Answer 2

Hey c:
So where we stuck?
Tried taking a derivative yet?

Answer 3

Doing it right now, I got scared because of the wording of the problem but I think it's actually pretty simple. Hold on! :)

Answer 4

Taking the deriv

Answer 5

With a problem like this can I assume it's with respect to x or is that a nono?

Answer 6

Yes, that's fine :)
Take derivative with respect to x.

We assume y is y(x), a function of x, even though it's not written explicitly.

Answer 7

<3 I'll get back to ya if I get stuck!

Answer 8

k np.
boy that derivative is going to be a doozy lol
especially on the right hand side

Answer 9

cos(xy-2)*y'+1=(y'(cosx-2y)+(y-1)(-sinx-2)*y'

Did I screw it up?

Answer 10

Extra ( whoops

Answer 11

Both sides look a tiny bit off.. let's start with the left side.

Answer 12

Ohh I needed mult rule on left I think

Answer 13

@tjb69812 may i help u

Answer 14

\[\Large\rm \cos(xy-2)\color{royalblue}{(xy-2)'}+1\]

Answer 15

Yah chain rule for this first term in the blue :)

Answer 16

1(y)+x(y')

Answer 17

sorry product rule* yes what you said :)

Answer 18

mm looks good!

Answer 19

\[\Large\rm \cos(xy-2)\color{orangered}{(y+xy')}+1\]

Answer 20

On the right side:\[\Large\rm (y-1)\cos(x-2y)\]Product rule again, looks like you got that started correctly,\[\Large\rm \color{royalblue}{(y-1)'}\cos(x-2y)+(y-1)\color{royalblue}{(\cos(x-2y))'}\]And then I think you must made a small error on the end there

Answer 21

y'*cos(x-2y)+(y-1)(-sin(x-2y)*[(1-2)*y']

???

Answer 22

Hahah it's late I may be goofing up again

Answer 23

mmm that looks closer:\[\large\rm \color{orangered}{(y')}\cos(x-2y)+(y-1)\color{orangered}{(-\sin(x-2y))\color{royalblue}{(x-2y)'}}\]
\[\large\rm \color{orangered}{(y')}\cos(x-2y)+(y-1)\color{orangered}{(-\sin(x-2y))\color{orangered}{(1-2y')}}\]

Answer 24

That last y' should be inside the brackets, yah?

Answer 25

Yessir you're correct on that

Answer 26

So we've got something like this:
\[\rm \cos(xy-2)(y+xy')+1=(y')\cos(x-2y)+(y-1)(-\sin(x-2y))(1-2y')\]

Answer 27

Time to plug in?

Answer 28

Or wait isolate y' right

Answer 29

We COULD solve for y' and see that it's a function involving both x and y.
So we'll think of y' as \(\Large\rm y'(x,y)\).

BUT solving for y' is soooo much work here.
Plugging in first will simplify things quite a bit I think.

Answer 30

cos(2*1*-2)*(1+2*y')+1 = y' * cos(2-2*1)

Answer 31

Doesn't the last term just = 0 that's what took me a second hahah

Answer 32

Yea nvm it does

Answer 33

\[\Large\rm \cos(2-2)(1+2y')=y' \cos(2-2)+(1-1)...\]
Yah we can totally ignore that last guy :)

Answer 34

\[\Large\rm \cos0(1+2y')=y' \cos0\]

Answer 35

Shouldn't be too hard to solve from there, just a few algebra steps, yah?

Answer 36

Safe to plug in 1 for cos0?

Answer 37

Yup, that looks accurate!

Answer 38

\[\Large\rm 1+2y'=y'\]

Answer 39

I think we dropped a 1 somewhere or am I seeing things

Answer 40

(1+2*y')+1 on the end

Answer 41

On the left side? Ooo good call

Answer 42

my beeeaaddd,\[\Large\rm \cos(2-2)(1+2y')+1=y' \cos(2-2)+(1-1)...\]\[\Large\rm \cos0(1+2y')+1=y'\cos0\]\[\Large\rm 1(1+2y')+1=y'\cdot 1\]\[\Large\rm 1+2y'+1=y'\]

Answer 43

I'm getting y' = -2 after that yeah

Answer 44

so y-1=-2(x-2) :D

Answer 45

So y' gave you your slope? Good good good?
Writing it in point-slope form? Yah that seems like the best idea here :) Since we're given a point.


Yayyy team! \c:/ Good job!

Answer 46

Thanks a ton dude!

Answer 47

np