If y= cos(sin X). d²y/dx²+Tanx.dy/dx+ycos²x=?

2 Answers
Mar 30, 2018

Please see below.

Explanation:

Here,

#color(red)(y=cos(sinx).......................to(I)#

Diff.w.r.t.x , using Chain Rule

#(dy)/(dx)=-sin(sinx)d/(dx)(sinx)#

#color(red)((dy)/(dx)=-sin(sinx)cosx....to(II)#

Again diff.w.r.t.x, using Product and Chain Rule,

#(d^2y)/(dx^2)=-[sin(sinx)d/(dx)(cosx)+cosxd/(dx)(sin(sinx)) ]#

#(d^2y)/(dx^2)=-[sin(sinx)(-sinx)+cosxcos(sinx)cosx]#

#(d^2y)/(dx^2)=sin(sinx)(sinx)-cos(sinx)cos^2x]#

#(d^2y)/(dx^2)=sin(sinx)color(blue)cosx(sinx/color(blue)cosx)- cos(sinx)cos^2x#

Using #(I) and(II),#we get

#(d^2y)/(dx^2)=(-(dy)/(dx))(tanx)-ycos^2x#

#(d^2y)/(dx^2)+tanx(dy)/(dx)+ycos^2x=0#

Mar 30, 2018

#(d^2y)/(dx^2)+tanx(dy)/(dx)+ycos^2x=0#

Explanation:

As #y=cos(sinx)#

#(dy)/(dx)=-sin(sinx)*cosx# i.e. #sin(sinx)=-1/cosx(dy)/(dx)#

and using product formula

#(d^2y)/(dx^2)=-cosxcos(sinx)cosx+sin(sinx)*sinx#

= #-cos^2xcos(sinx)+sinxsin(sinx)#

= #-cos^2x*y+sinx*(-1/cosx(dy)/(dx))#

= #-ycos^2x-tanx(dy)/(dx)#

and hence #(d^2y)/(dx^2)+tanx(dy)/(dx)+ycos^2x=0#