how would you implicit differentiate this?
x + y = tan(x - y)
I can't isolate y'
thanks
the square is on the secant, not the (x - y)
$\displaystyle 1 + y' = \sec^2 (x - y) (1 - y')$
you don't have many options here, what can we do with this. there are brackets on the right, multiply them out, we get:
$\displaystyle 1 + y' = \sec^2 (x - y) - y' \sec^2 (x - y)$
we want to solve for $\displaystyle y'$, so get all the temrs with $\displaystyle y'$ on one side of the equation
$\displaystyle \Rightarrow y' + y' \sec^2 (x - y) = \sec^2 (x - y) - 1 $
$\displaystyle y'$ is a common term on the left, factor it out:
$\displaystyle y' \left( 1 + \sec^2 (x - y) \right) = \sec^2 (x - y) - 1$
divide both sides by $\displaystyle 1 + \sec^2 (x - y)$:
$\displaystyle \Rightarrow y' = \frac {\sec^2 (x - y) - 1}{\sec^2 (x - y) + 1}$
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:
$\displaystyle \Rightarrow y' = \frac {\tan^2 (x - y)}{\sec^2 (x - y) + 1}$