Question

How do you find `(dy)/(dx)` given `xe^siny=e^y`?

Answer 1

`rArr(dy)/(dx)=(-e^siny)/(cosye^sinyx-e^y)`

Explanation:

The implicit differentiation of the given equation is determined by applying some differentiation properties

`color(blue)((u+v)'=(du)/dxxxv+(dv)/dxxxu`

`color(red)((de^u)/dx=(du)/dxxxe^u)`

`color(brown)((dsiny)/(dx)=(dy)/(dx)xxcosy)`

Start computing `(dy)/(dx)`

`xe^(siny)=e^y`

`rArrd/dx(xe^(siny))=d/dx(e^y)`

`rArrcolor(blue)(dx/dxxx(e^(siny))+((de^siny)/dx)xxx)=color(red)(dy/dxxxe^y)`

`rArre^siny+color(red)((dsiny)/dx)xxe^sinyxxx=dy/dxxxe^y`

`rArre^siny+color(red)((dsiny)/dx)xxe^sinyxxx=dy/dxxxe^y`

`rArre^siny+color(brown)((dy)/(dx)(cosy))e^sinyx=dy/dxe^y`

`rArr(dy)/(dx)(cosy)e^sinyx-dy/dxe^y=-e^siny`

`rArr(dy)/(dx)(cosye^sinyx-e^y)=-e^siny`

`rArr(dy)/(dx)=(-e^siny)/(cosye^sinyx-e^y)`

Answer 2

`xe^(siny)=e^y`

`e^(lnx)e^(siny)=e^y`

`e^(lnx+siny)=e^y`

`lnx+siny=y`

Use use normal and implicit differentiation...

`1/x+cosy*(dy)/(dx)=(dy)/(dx)`

`(dy)/(dx)-cosy*(dy)/(dx)=1/x`

`(dy)/(dx)(1-cosy)=1/x`

`(dy)/(dx)=1/(x*(1-cosy))`

Answer 3

`dy/dx = 1/(x(1 - cosy))`

Explanation:

I would solve the problem by taking natural logarithms;

`xe^siny=e^y`
`:. ln(xe^siny) = ln(e^y)`
`:. lnx + ln(e^siny) = lne^y`
`:. lnx + siny = y`

Differentiate wrt `x`
`:. d/dxlnx + d/dxsiny =d/dx y`

`:. 1/x + d/dysinydy/dx = dy/dx`

`:. 1/x + cosydy/dx = dy/dx`

`:. dy/dx - cosydy/dx = 1/x`

`:. dy/dx(1 - cosy) = 1/x`

`:. dy/dx = 1/(x(1 - cosy))`

Answer 4

`1/(x(1-cos y))`

Explanation:

This can be organized as

`x = e^y/(e(sin y))=e^(y-sin y)`. So,

`y'=1/((dx)/(dy))`

`=1/(e^(y-sin y)(1-cos y))`

`=1/(x(1-cos y)`,