Find dy/dx of x cos y= sin (x+y) ?

3 Answers
Mar 9, 2018

# dy/dx=(xtanx+1-secx)secxcos^2y#.

Explanation:

We will use the Usual & the Rule of Implicit Differentiation.

#xcosy=sin(x+y)=sinxcosy+cosxsiny#,

#:. (x-sinx)cosy=cosxsiny#.

Dividing by #cosx*cosy#, we get,

# (x-sinx)/cosx=siny/cosy#.

#:. x/cosx-sinx/cosx=siny/cosy, i.e., #

#tany=xsecx-tanx#.

#:. d/dx(tany)=d/dx(xsecx-tanx)#.

#:. sec^2y*dy/dx=x(secxtanx)+1*secx-sec^2x#,

#=(xtanx+1-secx)secx#.

# rArr dy/dx={(xtanx+1-secx)secx}/sec^2y, or, #

# dy/dx=(xtanx+1-secx)secxcos^2y#.

Enjoy Maths.!

Mar 9, 2018

This type of question must be of the form :"If #xcosy=sin(x+y)#,then prove that #(dy)/(dx)=(given)#.otherwise there are different answers.
#(dy)/(dx)=(cosx+xsinx-1)/(x^2-2xsinx+1)#

Explanation:

#color(blue)(d/(dx)(tan^(-1)X)=1/(1+X^2)#
#xcosy=sin(x+y)=sinxcosy+cosxsiny#
#=>xcosy-sinxcosy=cosxsiny#,
dividing both sides by #cosy!=0,# we get
#x-sinx=cosxtany=>tany=(x-sinx)/cosx#
#=>y=tan^(-1)((x-sinx)/cosx)#, Let, #color(red)(u=(x-sinx)/cosx)#, then
#(dy)/(dx)=1/(1+((x-sinx)/cosx)^2)(cosx(1-cosx)-(x-sinx)(-sinx))/(cos^2x)#
#=cancel((cos^2x))/(cos^2x+(x-sinx)^2)*(cosx-cos^2x+xsinx-sin^2x)/cancel((cos^2x))#
#=(cosx+xsinx-color(red)((cos^2x+sin^2x)))/(color(red)(cos^2x)+x^2-2xsinx+color(red)(sin^2x))#
#=(cosx+xsinx-1)/(x^2-2xsinx+1)#

Mar 9, 2018

Image reference...

Explanation:

My notebook...

Hope it helps....
Thank you...