Find dy/dx if?

x^cosy+y^sinx=1

2 Answers
Apr 13, 2018

(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)

Apr 13, 2018

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.