IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - Differentiate-

BuddyGuy

Find dy/dx
y=cos^-1[(3+5cosx)/(5+3cosx)]

vinu

cos y = [(3+5cosx)/(5+3cosx)] ;

differenciate......

- sin y dy/dx = [(5+3cosx)(-5sinx) - (3+5cosx)(-3sinx)] / (5+3cosx) ²= [-16sinx]/ (5+3cosx) ²;

transform siny 2 cosy & simplify....

dy/dx = 4 / (5+3cosx) .

Sree

Taking cos on both sides
cos y = cos ( cos^-1 (( 3+5cos x ) / ( 5+3cos x )))
cosy = ( 3+5cos x ) / ( 5+3cos x )
differentiating on both sides
( -sin y )( dy/dx ) = (( 5+3cos x )( 0+5(-sin x )) - ( 3+5cos x )( 0+3(-sin x )))/( 5+3cos x )^2.
( -sin y )( dy/dx ) = ( -16sin x ) / ( 5+3cos x )^2.
Therefore , ( dy/dx ) = ( 16sinx ) / ( (sin y) ( 5+3cos x )^2 )

a_buddy4uin06

jus temme if ter is any error in this..

cosy = (3 + 5cosx)/ (5 + 3cosx)

cross multiplyin,
5cosy + 3cosxcosy = 3 +5cosx

diff wrt x
-5sinydy/dx -3sinxcosy - 3cosxsinydy/dx = -5sinx

hence dy/dx= sinx(5 - 3cosy)/siny(5 + 3cosx)

amar.gupta

good work by almost every body especially vinu.

but if you calculate the value of siny it will be much simpler.

sin y dy/dx = [16sinx]/ (5+3cosx) ²

and cos y = [(3+5cosx)/(5+3cosx)]

so sin^2y = 1 -[(3+5cosx)/(5+3cosx)] ^2

or sin^2y = [(5+3cosx)^2 -(3+5cosx)^2]/(5+3cosx) ^2

or sin^2y = [(8+8cosx)(2-2cosx)]/(5+3cosx) ^2

or sin^2y = 16(1+cosx)(1-cosx)/(5+3cosx) ^2

or sin^2y = 16sin^2x/(5+3cosx) ^2

or siny = 4sinx/(5+3cosx)

now [4sinx/(5+3cosx)] dy/dx = [16sinx]/ (5+3cosx)^2

or dy/dx = 4/ (5+3cosx)

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