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differentiating inverse trig

Hi im stuck on this question

given that x and y satisfy the equation tan^-1 x + tan^-1 y + tan^-1 (xy)=11/12 pi
prove that when x=1 dy/dx= -1-rt3/2

i worked out when x=1 y=rt3 but i think im differentating wrong because i get the wrong answer! im a bit unsure how to do the tan^-1 (xy) term

thanks for any help!
differential of arctan x is 1/(1+x^2)
Differentiate tan^-1(xy) with respect to x gives you y/(1+(xy)^2)
sarah12345
Hi im stuck on this question

given that x and y satisfy the equation tan^-1 x + tan^-1 y + tan^-1 (xy)=11/12 pi
prove that when x=1 dy/dx= -1-rt3/2

i worked out when x=1 y=rt3 but i think im differentating wrong because i get the wrong answer! im a bit unsure how to do the tan^-1 (xy) term

thanks for any help!

tan^-1x diffs to 1/x^2+1
tan^-1y diffs to 1/y^2+1
tan^(-1)(xy)
let u=xy
d(tan^(-1)(xy))/dx=du/dx.d(tan^-1u)/du
=[y+xdy/dx]/1+(xy)^2

so
1/x^2+1+1/y^2+1dy/dx+[y+xdy/dx]/1+(xy)^2=0
when x=1 y=rt(3)
1/2+1/4dy/dx+[rt(3)+dy/dx]/4=0
2+rt(3)+dy/dx[1+rt(3)]=0
dy/dx=[-2-rt(3)]/[1+rt(3)]
=[(1-rt(3))*(-2-rt(3)]/-2
=[{1-rt(3)}*(2+rt(3)]/2
=[-1-rt(3)]/2
Reply 4
Avatar for sarah12345
sarah12345
OP
2
thanks for the help but i dont understand this bit

let u=xy
d(tan^(-1)(xy))/dx=du/dx.d(tan^-1u)/du
=[y+xdy/dx]/1+(xy)^2


i understand that dy/dx=du/dx * dy/du

so why doesnt this give
y/(1+(xy)^2) ??
sarah12345
thanks for the help but i dont understand this bit

let u=xy
d(tan^(-1)(xy))/dx=du/dx.d(tan^-1u)/du
=[y+xdy/dx]/1+(xy)^2


i understand that dy/dx=du/dx * dy/du

so why doesnt this give
y/(1+(xy)^2) ??

u=tan^-1 u
so du/dx =1/(1+u^2)=1/(1+(xy)^2)
now imagine you had 1=xy and want to diff with respect to x
this would give you y+xdy/dx=0
well for u=xy the u diffs to du/dx so
du/dx=y+xdy/dx
mult the two together we get

1/(1+(xy)^2).y+xdy/dx
Reply 6
Avatar for sarah12345
sarah12345
OP
2
o ok thanks alot!

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