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

Question

Elijah Abram · Accepted Answer

# 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)}.# 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.!

William Abdalla · Answer

undefined To differentiate the given expression, apply the chain rule and the product rule:

1. Differentiate $ y \ln(e^{x-y^3}) $:
$$ \frac{d}{dx}[y \ln(e^{x-y^3})] = y \frac{d}{dx}[x - y^3] + \ln(e^{x-y^3}) \frac{dy}{dx} $$

2. Differentiate $ x e^{x-y} $:
$$ \frac{d}{dx}[x e^{x-y}] = e^{x-y} + x \frac{d}{dx}[e^{x-y}] $$

3. Combine the results from steps 1 and 2 to find the overall derivative.