IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

iitaspirant10

If cosy=xcos(a+y) ,

Prove: dy/dx= [cos²(a+y)] / [sina]

iitaspirant10

pls reply

studyid

hello iitaspirant10 ...

see.. here is my solution....

cosy = xcos(a+y)

diff. both sides w.r.t. x,

-siny. y' = -xsin(a+y).y' + cos(a+y)

therefore.. y' = cos(a+y)/(xsin(a+y)-siny)

then take the value of x from the given ques. put it in the above... take lcm and you will get your answer...

and ya rate me... i u think ...... that the sol. is understood to u..

Harsh137

cos y = x cos (a + y) => x = cos y/ cos(a+y)
Differentiating w.r.t. x

-sin y dy/dx = x*(-sin(a+y)*dy/dx) + cos(a+y)*dx/dx
-sin y y'= -x sin(a+y) y' + cos(a+y) ........[dy/dx=y']
y'(x sin(a+y) - sin y)=cos(a+y)
y'= cos (a+y)/(x sin(a+y) - sin y)
=cos(a+y)/[{cos y sin(a+y)/cos(a+y)} - sin y]
=cos²(a+y)/[cos y sin(a+y) - sin y cos(a+y)]
=cos²(a+y)/sin(a+y-y)

dy/dx= [cos²(a+y)] / [sina]

mukulss

we can do this que explicitly and implicitly,
both ways kkkkkkkkk:

M1:

explicitly:

cos y = xcos(a+y)

1 = x[cosa - sinatany]

implies,
solving on more ,
u get:

sina* tany = cosa - 1/x

tany = cota - 1/xsina

this way u get an explaicit relation between x and y:

now diferentiating:

sec^2 y dy/dx = 1/x^2 sina

dy/dx = [cos2(a+y)] / [sina]

mukulss

M2:[very simple and very food too]

by making use of an implicit relation:

see in this type:

use:

dx/dy = [cos(a+y)][-siny]+ [cosy][sin(a+y)]
--------------------------------------------
[cos2(a+y)]
see now:

dx/dy = sina / [cos(a+y)]^2

mukulss

hey question putter,

what abt my both answerts,,,,,,,,,,,,,

mukulss

can nyone explain who rated me 4:
"""""""""
hey question putter,

what abt my both answerts,,,,,,,,,,,,, """"""""""""""""""""""""""

Ask & Discuss Questions with Community & Experts