Question

Really stumped!
Differentiate implicitly: tan(x+y)=(x-y)

Answer 1

Okay... let's get to work :)
First, pretend that y = y(x)
in that y is a function of x. Then differentiate normally :)
For instance, at the right first, because it's easier, what's the derivative (with respect to x) of x-y ?

Answer 2

So would I have y'-y(x) then?

Answer 3

hmm...
no :)
Let's differentiate x - y slowly...
What's the derivative of x?

Answer 4

It's 1, isn't it? And then the derivative of y is just y'
So the derivative of x - y is just 1 - y'
Catch me so far?

Answer 5

I think so. Our prof tried to cover this section with about 7 mins left in class before spring break. So, this is quite blurry to me :)

Answer 6

Okay, so the derivative of the right side is 1 - y', that much is clear, aye? :)
Now let's get to work on the left side.
\[\large \frac{d}{dx}\tan(x+y)\]

Answer 7

So would the left side then be d/dx tan(x+y) + d/dx (x) + d/dx (y)?

Answer 8

You...seem to know much more about this than you'd reveal. Yes, in fact, this is true :)
Let me just derive it for you :)
\[\large \frac{d}{dx}\tan(x+y)=\sec^2(x+y)\cdot\frac{d}{dx}(x+y)\]
Can you do the rest?

Answer 9

This is just the chain rule, no?

Answer 10

Oh that chain rule :)

Answer 11

So,you understand how this occurred?
\[\large \frac{d}{dx}\tan(x+y)=\sec^2(x+y)\cdot\frac{d}{dx}(x+y)\]
?

Answer 12

Yes, I do understand how you came to this.

Answer 13

So just perform the rest... there's the (d/dx)(x+y) to worry about.

Answer 14

So I need my y' on one side. So if I did this correct then, I shouild have 2y'=1-sec^2(x+y)

Answer 15

Oh maybe not.

Answer 16

Slowly...
What's
\[\frac{d}{dx}(x+y) \ \ ?\]

Answer 17

Forgot that :S

Answer 18

You just distribute the differentiation operator.

Answer 19

So then d/dx (x+y) = 1+ y'?

Answer 20

That's right :)

Answer 21

And I thought I was confusing mysfelf here.

Answer 22

So, now we have...
\[\huge \sec^2(x+y)\cdot(1+y')=1-y'\]
Now just solve for y'

Answer 23

Yes! That is what I have!!

Answer 24

Solve for y' correct?

Answer 25

Yes.

Answer 26

Ok, that wasn't as difficult as I was thinking it was. Thanks for your help!

Answer 27

No problem :)