if x^siny + y^sinx =1 ,then find  dy/dx

Differentiating both sides of the given equation we get :

(siny).(x^{siny - 1}) + (x^siny)(logsiny)(cosy.dy/dx) + (sinx)(y^{sinx - 1})(dy/dx) + (y^sinx)(logsinx)(cosx) = 0

After simple rearrangement you can easily get dy/dx.

Solving by using log will be slightly easy:

Taking log on both sides, you get,

(sin y)( log x) + (sin x)( log y)=0

Now differentiate wrt x,

you will get,

{(sin y)/x+(log x)(cos y)(dy/dx)} +( -cos x)/y(dy/dx)+(log y)(cos x)=0

Simplify this and u'll surely get the answer.

X^SinY + Y^SinX = 1

dY/dX = ?

Let  U = X^SinY  & V =  Y^SinX

Thus U + V = 1

or dU/dX + dV/dX = 0            ...(1)

Now From U = X^SinY  

ln U = SinY ln X

or (1/U)dU/dX = (CosY.ln X)dY/dX + (SinY)/X

or dU/dX = U (CosY.ln X)dY/dX + U (SinY)/X

or dU/dX = X^SinY (CosY.ln X)dY/dX + X^SinY (SinY)/X

or dU/dX = X^SinY (ln X^CosY )dY/dX + X^SinY-1.(SinY)                 ...(2)

Also From V =  Y^SinX

or ln V = SinX .lnY

or (1/V)dV/dX = CosX .lnY + (SinX /Y) dY/dX

or dV/dX = V.CosX .lnY + V.(SinX /Y) dY/dX

or dV/dX = Y^SinX.CosX .lnY + Y^SinX.(SinX /Y) dY/dX

or dV/dX = Y^SinX. lnY^CosX + Y^SinX-1. SinX .dY/dX                    ...(3)

Form equations (1), (2) and (3)

dU/dX + dV/dX = 0

or X^SinY (ln X^CosY )dY/dX + X^SinY-1.(SinY)+ Y^SinX. lnY^CosX + Y^SinX-1. SinX .dY/dX       = 0

or dY/dX (X^SinY .ln X^CosY + Y^SinX-1. SinX) = - [ X^SinY-1.(SinY)+ Y^SinX. lnY^CosX ]

or dY/dX = - [ X^SinY-1.(SinY)+ Y^SinX. lnY^CosX ] /[X^SinY .ln X^CosY + Y^SinX-1. SinX]

or dY/dX = - [ X^SinY-1.(SinY)+ Y^SinX. lnY^CosX ] /[ Y^SinX-1. SinX  + X^SinY .ln X^CosY ]

A correction in my above post:

Plz correct this step:

{(sin y)/x+(log x)(cos y)(dy/dx)} +( -cos x)/y(dy/dx)+(log y)(cos x)=0

The corrected step is:

{(sin y)/x+(log x)(cos y)(dy/dx)} +( sin x)/y(dy/dx)+(log y)(cos x)=0