Find y' in these 2 equations?

#x.e^(-y/2)##+ y.e^(-x/2)##=sin(x^2 + y^2)#
#and #

# y=x^(lnx) . (sec x)^(3x)#

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2 Answers
maganbhai P.
Aug 12, 2018

#y'=(4xcos(x^2+y^2)-2e^(-y/2)+ye^(-x/2))/(2e^(-x/2)-xe^(-y/2)-4ycos(x^2+y^2))#

Explanation:

Let , #y'=dy/dx=y_1#

Here ,

#(1)x*e^(-y/2)+y*e^(-x/2)=sin(x^2+y^2)#

Diff. w.r.t. # x #,using product and chain rules :

#xe^(-y/2)(-1/2)y_1+e^(-y/2)+ye^(-x/2)(-1/2)+e^(-x/2) *y_1#

#color(white)(......................................)=cos(x^2+y^2)[2x+2yy_1]#

#:.xe^(-y/2)(-1/2)y_1+e^(-x/2) *y_1-cos(x^2+y^2)2yy_1#

#color(white)(...........................)=2xcos(x^2+y^2)-e^(-y/2)-ye^(-x/2)(-1/2)#

#y_1{e^(-x/2)-x/2e^(-y/2)-2ycos(x^2+y^2)}#

#color(white)(...........................)=2xcos(x^2+y^2)-e^(-y/2)+y/2e^(-x/2)#

#y_1{2e^(-x/2)-xe^(-y/2)-4ycos(x^2+y^2)}#

#color(white)(...........................)=4xcos(x^2+y^2)-2e^(-y/2)+ye^(-x/2)#

#:.y_1=(4xcos(x^2+y^2)-2e^(-y/2)+ye^(-x/2))/(2e^(-x/2)-xe^(-y/2)-4ycos(x^2+y^2))#

For second eqn. ,please see next answer.

maganbhai P.
Aug 12, 2018

#(dy)/(dx)=x^(2lnx-1)*lnx^2+3(secx)^(6x)[xtanx+ln(secx)]#

Explanation:

Let , #y'=(dy)/(dx)=y_1#

#(2)y=x^lnx*(secx)^(3x)=u*v.......................(i)#

Let ,

#u=x^lnx# ,#..................and v=(secx)^(3x)#

Taking natural log both sides we get

#lnu=ln(x^lnx)# , #................andlnv=ln((secx)^(3x))#

#:.lnu=lnx*lnx =(lnx)^2 ............and lnv=3xln(secx)#

Diff.w.r.t. #x#

#1/u(du)/(dx)=2lnx*1/xand 1/v(dv)/(dx)#=#3x*(secxtanx)/secx+ln(secx)*3#

#1/u(du)/(dx)=1/xlnx^2............and 1/v(dv)/(dx)#=#3xtanx+3ln(secx)#

#(du)/(dx)=u/xlnx^2............and (dv)/(dx)=3v[xtanx+ln(secx)]....(ii)#

Now ,

#y=u*vto "Product Rule" #

#(dy)/(dx)=u(dv)/(dx)+v(du)/(dx)#

From #(i) and (ii)#

#:.(dy)/(dx)=x^lnx [u/xlnx^2]+(secx)^(3x)[3v(xtanx+ln(secx))]#

Subst. back values of #u and v#
#(dy)/(dx)#=#x^lnx [x^lnx/xlnx^2]+(secx)^(3x)[3(secx)^(3x)(xtanx+ln(secx))]#

#(dy)/(dx)=x^(2lnx-1)*lnx^2+3(secx)^(6x)[xtanx+ln(secx)]#

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