IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

zeeshanp

sin(x+y)=log(x+y)

RyuAmakusa

put x+y=t now dy/dx +1 = dt/dx

and sin t =log t so t=e^sin t . now diff.

dt/dx = e^sin t.* cost * dt/dx. ......

now cant u get it.

mohile

sin(x+y)=log(x+y) then sinxcosy+cosxsiny=log(x+y) then dy/dx(cosxcosy-sinxsiny)+(cosxcosy-sinxsiny)=1/x+y(1+dy/dx) ,then [cos(x+y)-1/x+y]dy/dx=1/x+y-cos(x+y) then dy/dx= -[1-cos(x+y)(x+y)/1-cos(x+y)(x+y)] then dy/dx= -1

ganesha1991

differentiate both sides
cos(x+y) [ 1+dy/dx] = 1/(x+y) [ 1+ dy/dx]
cos(x+y) + cos(x+y) dy/dx = [1+dy/dx/ (x+y)
(x+y) cos(x+y) + (x+y) cos(x+y) dy/dx = 1 + dy/dx
dy/dx [ (x+y) cos(x+y) -1] = 1 - (x+y) cos(x+y)
dy/dx = - [ (x+y)cos(x+y) -1] / [(x+y) cos(x+y) -1]
= - 1

varshavallig

simple.....

sin(x+y)-log(x+y)=0
diff, cos(x+y) * (1+y') - (1/(x+y) )*(1+y')=0
(1+y')[cos(x+y)-(1/(x+y))=0
y'=-1

!!!!!!!!!!!!!!!!!!!!!

siva7792

Sin(x+y)=log(x+y)
Sin(x+y)-log(x+y)=0
Cos(x+y)[1+dy/dx]-1/(x+y)[1+dy/dx]=0
[1+dy/dx]{cos(x+y)-1/(x+y)}=0
==1+dy/dx=0
=dy/dx=-1

y`=-1

rate me

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