How do I find dy/dx by implicit differentiation?

Question

Find $\displaystyle \frac{dy}{dx}$ by implicit differentiation.

$\tan(x+y)=x$

So far, I got to

$\displaystyle \sec^2(x+y) \left(1+\frac{dy}{dx}\right) = 1$

but then im lost.. can someone please help and explain? I would really appreciate it!!

Answer 1

Well, you have $\sec^2(x+y) (1+ y') = 1$, from which you get $y' = \cos^2(x+y)-1$.

Answer 2

you need to foil sec$^2(x+y)(1+ y'(x)) =1$ and then solve for $y'(x)$.

Think about sec$^2(x+y)$ as one whole.

Hint: the answer will be in terms of $x$ and $y$