implicit differentiation question. help pls?

Question

is the implicit differentiation os:
sec(x+y) = tanx -tany..
show work:
sec(x+y)tan(x+y)*y' = sec^2x-sec^2y*y'
let sec(x+y)tan(x+y) = A
...

so is..
y' = sec^2x / A+sec^2y

i have no idea, if that is right.
if it's wrong would you please show me how.
thanks :)

you didnt even read my question first poster. i did do it, i was asking for somebody to check the answer.

Answer 1

You have one error. In the derivative d/dx sec(x+y): Let u = x+y, then derivative of d sec(u)/dx = tan(u)*sec(u)*du/dx. The implicit derivative du/dx is d(x+y)/dx which is 1 + y'. You left out the "1".

Then you want to get all the y' to one side of the equation if possible. Using your substitution of A=sec(x+y)tan(x+y) the equation is

A + A*y' = sec^2(x) - sec^2(y)*y'

A*y' + sec^2(y)*y' = sec^2(x) - A

y' = [sec^2(x) - A]/[sec^2(y) + A]

You should then re-insert your value for A to complete the equation
EDIT: You should also try to simplify using trig identities if possible.

Answer 2

sec (x+y) = tan x - tan y
sec (x+y) tan (x+y) (dx+dy) = sec² x dx - sec² y dy
sec (x+y) tan (x+y) (1+dy/dx) = sec² x - sec² y dy/dx
sec (x+y) tan (x+y) + sec (x+y) tan (x+y) y' = sec² x - sec² y y'
sec² x - sec (x+y) tan (x+y) = (sec² y + sec (x+y) tan (x+y)) y'
y'= (sec² x - sec (x+y) tan (x+y))/(sec² y + sec (x+y) tan (x+y))

Your derivation is almost right, you just misapplied it when finding the derivative of x+y - this is 1+y', not just y' (since the x doesn't disappear when differentiating w.r.t. x). It's easy to make these kinds of mistakes, which is why I got in the habit of explicitly writing out the dx and dy and then only dividing by dx at the very end of the differentiation.

Answer 3

No - DO YOUR OWN HOMEWORK - that's what it's for. HOME - WORK