Question

can anyone give me hint
if sec[(x+y)/(x-y)] =a
prove that dy/dx=y/x

should i substitute x=ycos2theta ???

Answer 1

or should i subtitute x=tantheta

Answer 2

yes!
x+y)/(x-y = sec^-1 a
and since 'a' is constant, the entire right side is constant and its derivative = 0 !

Answer 3

oh :OOOOOO
i don't need to subtitute anything :OOO

Answer 4

you can simplify it even more
let sec^-1 a = c (just for sake of simplicity)
(x+y)/ (x-y) = c
now do you know about componendo -dividendo ???

Answer 5

no :'( forgot

Answer 6

okk...
\(\large \dfrac{a}{b}=\dfrac{c}{d} \implies \dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
so, what about \(\large \dfrac{x+y}{x-y}=\dfrac{c}{1}\)
??

Answer 7

yo got it XD
saw it on the net :)

Answer 8

x+y+x-y /(x+y-x+y)
=2x/2y
=x/y :D

Answer 9

lol that is just
\(\dfrac{x}{y}=\dfrac{c+1}{c-1}\)
we have still even did not start differentiating!

Answer 10

\(\dfrac{x}{y}=\dfrac{c+1}{c-1}=k\)
k is another constant for simplicity ...
now differentiate left of x/y = k using quotient rule!
and since k is constant, right side = 0

Answer 11

@kittycat01 , still here ?

Answer 12

oh, dear! she is gone without realizing that she had not completed the solution (didn't even start differentiating!) :P

Answer 13

(x/y)' how? partial derivative or y respect to x? @hartnn

Answer 14

not partial, implicit differentiation
diff. both sides w.r.t x

Answer 15

\(\huge \dfrac{y-x \dfrac{dy}{dx}}{y^2}=0\)

Answer 16

and the numerator =0 solve for dy/dx?

Answer 17

yup

Answer 18

Thanks.

Answer 19

@hartnn -lol sorry i got so excited that i closed my net and i start doing al the sums again