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

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 ?

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

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

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?

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 :)

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. \[\large \frac{d}{dx}\tan(x+y)\]

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

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?

This is just the chain rule, no?

Oh that chain rule :)

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

Yes, I do understand how you came to this.

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

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

Oh maybe not.

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

Forgot that :S

You just distribute the differentiation operator.

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

That's right :)

And I thought I was confusing mysfelf here.

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

Yes! That is what I have!!

Solve for y' correct?

Yes.

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

No problem :)

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