Find dy/dx if?

Here,

#x^cosy+y^sinx=1#

Let, #u+v=1,where,#

#u=x^cosy and v=y^sinx#

Taking log both sides,

#=>lnu=lnx^cosy and lnv=lny^sinx#

#=>lnu=cosylnx...to(1) and lnv=sinxlny...to(2)#

Diff#.to(1)w.r.t. x# #"using "color(blue)"Product Rule and Chain Rule"#

#1/u(du)/(dx)=cosyxx1/x+lnx (-siny)(dy)/(dx)#

#(du)/(dx)=x^cosy[cosy/x-sinylnx(dy)/(dx)]#

Now diff#.to(2)w.r.t.x# #"using "color(blue)"Product Rule and Chain Rule"#

#1/v(dv)/(dx)=sinx1/y(dy)/(dx)+lnycosx#

#(dv)/(dx)=y^sinx[sinx/y(dy)/(dx)+cosxlny]#

Hence, #u+v=1=>(du)/(dx)+(dv)/(dx)=0=>(du)/(dx)=-(dv)/(dx)#

#x^cosy[cosy/x-sinylnx(dy)/(dx)]=-y^sinx[sinx/y(dy)/(dx)+cosxlny]#

#{y^sinxsinx/y-x^cosysinylnx}(dy)/(dx)=-y^sinxcosxlny- x^cosycosy/x#

#(dy)/(dx)=-(y^sinxcosxlny+x^cosycosy/x)/(y^sinxsinx/y- x^cosysinylnx)#

if #u,v# are two functions of #x#
then #d/dxu^v=u^v*ln(u)*(dv)/dx+v*u^(v-1)*(du)/dx#

#color(red)"an easy way to memorize this rule is"##color(green)" to consider the function u to be a constant and differentiate it"# #color(blue)" then consider v to be a constant and differentiate it"#
to get both the first term and second term

so consider #x^(cosy)=z_1#
and #(y^sinx)=z_2#
so the upper formula will be #z_1+z_2=1#

differentiate

#z_1'+z_2'=0#

#z_1'=(-siny)(dy)/dx(lnx)x^cos(y)+cos(y)*x^((cosy)-1)*1#

#z_2'=(lny)y^sinxcosx+(sinx)y^((sinx)-1)dy/dx#

substitute for #z_1',z_2'#

#(-siny)*(dy)/dx(lnx)*x^cosy+cos(y)x^((cosy)-1)+(lny)y^sinxcosx+(sinx)y^((sinx)-1)dy/dx=0#

simplify
#dy/dx=-((cos(y)x^((cosy)-1)+(lny)y^sinxcosx))/((sinx)y^((sinx)-1)-x^cosy(siny)lnx#

I hope this was helpful.