Question

x cos y + y cos x = 1 at (0,1) implicit differentiation

Answer 1

dy/dx at (0,1)?

Answer 2

Seems reasonable. Show your best attempt peper! I'll show you where you need help!

Answer 3

An equation of the line tangent to the graph of x cosy + y cosx = 1 at the point (0,1)


So I think dy/dx at (0,1)

Answer 4

Yeah, show us your best attempt. :)

Answer 5

Me thinks you'll have to use Product Rule twice.

Answer 6

all I have is:
x(-siny dy/dx) + (cosy)(1) + y (-sinx) + (cosx)(dy/dx)

Answer 7

I like to think of implicit differentiation as really just normal differentiation you've done all along. For instance, let's look at:

y=x

Obviously, we know that y'=1 here. But a closer look we notice that:

\[y^1=x^1\]\[\frac{ d }{ dx }(y^1)=\frac{ d }{ dx }(x^1)\]\[1*y^{1-1}*\frac{ dy }{ dx }=1*x^{1-1}*\frac{ dx }{ dx }\]notice how I just did the normal power rule where I take the power, multiply it out front, subtract 1 from the exponent? Then I multiplied by the derivative of the inside function with the chain rule.

BUT I did exactly the same thing to both sides. However now we can clean this up a bit!

What is the 1-1 power on both y and x? 0! And anything to the zero power is just 1. So we can just leave ourselves with:
\[\frac{ dy }{ dx }=\frac{ dx }{ dx }\] And now, what about that dx/dx term? Well, how much does x change when you change x? Obviously there's a 1 to 1 correspondence going on here! So that too is 1.

\[\frac{ dy }{ dx }=1\]

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So now try this with something else, like \[y=x^2\] and if you reorganize it to look like this: \[y^{1/2}=x\] they will both have the same derivative!

Answer 8

okay...

Answer 9

A. (cos1)x+y=1
B. x+y=1
C.-(sin1)x+y=1
D.x-y=1
E. (tan1)x+y=1

Answer 10

Err, if you go back to your expanded derivative, you'll have to set it equal to 0, since you took the derivative of both sides of your original equation.
Then, try to get the terms that do not have (dy/dx) on the right side.
Then, factor out the (dy/dx) and bring the remains to the right side.
Now, you'll have an equation for dy/dx. Plug in the point.

Answer 11

I tried.. Twice in fact! And somehow managed to end up with 0 in the denominator.

Answer 12

Hmm, doing it quick in my head, Idk how you're getting 0 in the denominator if the point is (0,1).

Answer 13

Write out the equation for dy/dx again. I'm definitely not getting 0 in denominator.

Answer 14

Okay. So I do the product rule twice and get x(-siny dy/dx) + (cosy)(1) + y (-sinx) + (cosx)(dy/dx) = 0

so then I move things around and end up with:
dy/dx (-sinxy + cosx) = -coxy + sinxy

have I messed up in these steps?

Answer 15

On the left, you should have dy/dx (-siny*x + cosx)* ;D
Maybe that's where you messed up

Answer 16

You wrote y*-sinx, instead of x*-siny if you didn't see yet

Answer 17

what you said I should have on the left I do have.. don't I? I just failed to put the *

Answer 18

or where are you getting the y*-sinx instead of x*-siny

Answer 19

Look at what you wrote when you factored out the dy/dx.

Answer 20

(-sinyx = cosx) should be inside the parentheses

Answer 21

(-sinyx + cosx)*

Answer 22

on the left side?

Answer 23

Yeah

Answer 24

Now get dy/dx alone. You shouldn't get 0 in denominator now. You flipped the variables in one term, small mistake.

Answer 25

dy/dx (-sinxy + cosx) is what I had and you had (-sinyx + cosx)

can you explain to me how I am suppose to plug this in. I think I am misunderstanding, I had dy/dx = (-cosy +sinxy)/(-sinyx + cosx)

dy/dx = (-cos(1) + sin(0)(1)) / (-sin(1)(0) + cos(0))

dy/dx = (-cos1) / 0

soooooooo I have know idea, I guess, how or what I am supposed to do because I am obviously doing something incorrect.

Answer 26

Err, there's no 0 in the denominator silly! Look at the cosine again.

Answer 27

You get 0 for the first term, plus the cosine of 0...:OOO

Answer 28

.... and the cosine of 0 is 1?

Answer 29

Yes, yes it is. Congrats, that was almost painful.

Answer 30

A. (cos1)x+y=1
B. x+y=1
C.-(sin1)x+y=1
D.x-y=1
E. (tan1)x+y=1

sooooo my answer doesn't look like any of the optons... so thats painful haha

Answer 31

i like pain but i dont like implicit diff

Answer 32

options*

Answer 33

Implicit differentiation isn't hard when you know the Implicit Function Theorem ;)

Answer 34

it's not hard, i just heavily dislike it

Answer 35

Whoa, what. Those choices don't look right. This probably is gonna be painful. I didn't do the derivative yet.

Answer 36

I don't have time to go over it again :(

Answer 37

OHOHHHOH, you need to find a tangent line to the point.

Answer 38

I showed you the choices earlier! And that's alright, my exam will just kill me tomorrow.

Answer 39

Well, we just found the slope of the tangent line at the point. Use point-slope equation now. :D

Answer 40

What'd you get for dy/dx, first?

Answer 41

Oh...We did the hard part. We just gotta find the equation of the line. Algebra 1 stuff.

Answer 42

y - 1 = -cos(1)(x - 0)
y = -xcos(1) + 1

Answer 43

y + xcos(1) = 1, A. Good night.

Answer 44

You are the best!
Thank you so much for your help!

Answer 45

It'd be nice to say 'no prob,' but that was truly painful.

Answer 46

Good luck on your test.

Answer 47

That it was. My professor insists we do Purdue University problems because they are challenging. Thank you so much again! Hopefully it will be smooth sailing.

Answer 48

Ahh yes, Purdue has a good math/engineering department. Cheers.