How do you differentiate #-2=yln(e^(x-y^3))+xe^(x-y)#?

1 Answer
Jul 15, 2017

# dy/dx={xe^(x-y)+e^(x-y)+y}/{xe^(x-y)+4y^3-x}. or, #

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

Explanation:

Knowing that, #ln(e^(x-y^3))=(x-y^3)lne=x-y^3,# we have,

# -2=y(x-y^3)+xe^(x-y), or, xe^(x-y)=y(y^3-x)-2, i.e., #

# xe^(x-y)=y^4-xy-2.#

#:. d/dx{xe^(x-y)}=d/dx{y^4-xy-2}.#

Using the Product and Sum/Difference Rule, we get,

# xd/dx{e^(x-y)}+e^(x-y)d/dx{x}=d/dx(y^4)-d/dx(xy)-d/dx(2),# or,

# xd/dx{e^(x-y)}+e^(x-y)=d/dx(y^4)-{xd/dx(y)+yd/dx(x)}....(1)#

Here, by the Chain Rule, #d/dx{e^(x-y)}=e^(x-y)d/dx(x-y),#

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

Also, #d/dx(y^4)=d/dy(y^4)*dy/dx=4y^3dy/dx....................................(3).#

Utilising #(2) and (3)# in #(1),# we get,

#xe^(x-y)(1-dy/dx)+e^(x-y)=4y^3dy/dx-xdy/dx-y,#

# :. xe^(x-y)-xe^(x-y)dy/dx+e^(x-y)=(4y^3-x)dy/dx-y,#

#:. xe^(x-y)+e^(x-y)+y=xe^(x-y)dy/dx+(4y^3-x)dy/dx,#

#={xe^(x-y)+4y^3-x}dy/dx.#

#:. dy/dx={xe^(x-y)+e^(x-y)+y}/{xe^(x-y)+4y^3-x}.#

This Answer is quite fair, but it can be put as :

Using, #xe^(x-y)=y^4-xy-2, &, e^(x-y)=(y^4-xy-2)/x,#

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

#={x(y^4-xy-2)+(y^4-xy-2)+xy}/{x(y^4-xy-2+4y^3-x)},#

# rArr dy/dx={x(y^4-xy-2)+y^4-2}/{x(y^4-xy-2+4y^3-x)}.#

Enjoy Maths.!