Question

Find dy/dx if?

`x^cosy+y^sinx=1`

Answer 1

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

Explanation:

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)#

Answer 2

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

Explanation:

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.