How do you find #dy/dx# by implicit differentiation given #y^2=ln(2x+3y)#?

Question

3808 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-21T14:53:52

# dy/dx=2/(6y^2+4xy-3).#

#y^2=ln(2x+3y)#

#:. d/dx(y^2)=d/dx{ln(2x+3y)}.#

Using the Chain Rule, we have,

#d/dy(y^2)dy/dx=1/(2x+3y)d/dx(2x+3y).#

#:. 2ydy/dx=1/(2x+3y){2+3dy/dx}.#

#=2/(2x+3y)+3/(2x+3y)dy/dx.#

#:. {2y-3/(2x+3y)}dy/dx=2/(2x+3y).#

#:., if (2x+3y)ne0, {(4xy+6y^2-3)/cancel(2x+3y)}dy/dx=2/cancel(2x+3y).#

As, #(2x+3y)=0# makes #ln(2x+3y)# undefined, we have,

#:. dy/dx=2/(6y^2+4xy-3).#