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)