# Math Help - implicit help

1. ## implicit help

how would you implicit differentiate this?
x + y = tan(x - y)
I can't isolate y'
thanks

2. Originally Posted by cubs3205
how would you implicit differentiate this?
x + y = tan(x - y)
I can't isolate y'
thanks
what was the answer you got. tell us and we will show you how to isolate the y'

3. i simplified to
1 + y' = sec(x - y)^2(1 - y')

4. Originally Posted by cubs3205
i simplified to
1 + y' = sec(x - y)^2(1 - y')
the square is on the secant, not the (x - y)

$1 + y' = \sec^2 (x - y) (1 - y')$

you don't have many options here, what can we do with this. there are brackets on the right, multiply them out, we get:

$1 + y' = \sec^2 (x - y) - y' \sec^2 (x - y)$

we want to solve for $y'$, so get all the temrs with $y'$ on one side of the equation

$\Rightarrow y' + y' \sec^2 (x - y) = \sec^2 (x - y) - 1$

$y'$ is a common term on the left, factor it out:

$y' \left( 1 + \sec^2 (x - y) \right) = \sec^2 (x - y) - 1$

divide both sides by $1 + \sec^2 (x - y)$:

$\Rightarrow y' = \frac {\sec^2 (x - y) - 1}{\sec^2 (x - y) + 1}$

now this answer has a nice symmetry to it, but if you don't care about that, you can simplify the numerator on the right:

$\Rightarrow y' = \frac {\tan^2 (x - y)}{\sec^2 (x - y) + 1}$