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

4 Answers
Oct 29, 2016

#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)#

Oct 29, 2016

#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))#

Oct 29, 2016

#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))#

#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)#,