How do you find #dy/dx# by implicit differentiation given #y=e^(x+2y)#?

Question

3331 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-07T18:35:46

#dy/dx=e^(x+2y)/(1-2e^(x+2y))=y/(1-2y)#

#y=e^(x+2y)#

#:. dy/dx=e^(x+2y)d/dx(x+2y)#

#=e^(x+2y){1+2dy/dx}=e^(x+2y)+2e^(x+2y)dy/dx#

#:. dy/dx-2e^(x+2y)dy/dx=e^(x+2y)#

#:.{1-2e^(x+2y)}dy/dx=e^(x+2y)#

#:. dy/dx=e^(x+2y)/(1-2e^(x+2y))=y/(1-2y)#

As we have been asked to differentiate implicitly, we proceeded as above. Otherwise , we can use the following method :

#y=e^(x+2y) rArr lny=x+2y rArr lny-2y=x#

Now diff.ing both sides w.r.t. #y, 1/y-2=dx/dy, or (1-2y)/y=dx/dy#

#:. dy/dx=1/(dx/dy)=y/(1-2y)#

Enjoy Maths.!