How do you implicitly differentiate #-1=-y^2x+xy-ye^(xy) #?

Question

How do you implicitly differentiate #-1=-y^2x+xy-ye^(xy)   #?

Luke Alvarez · Accepted Answer

# dy/dx={y(1-y-ye^(xy))}/{2xy+xye^(xy)+e^(xy)-x},#

OR

# dy/dx={y^3(xy-x-1)}/(xy+xy^2+x^2y^2-x^2y^3+1).# To avoid #-ve#signs, let us rewrite the eqn. as, #y^2x+ye^(xy)=xy+1.# Diff.ing term-wise, using the usual Rules of Diffn., we get, #{y^2d/dx(x)+xd/dx(y^2)}+{yd/dxe^(xy)+e^(xy)d/dx(y)}# #=d/dx(xy)+0.# #:. y^2(1)+x{d/dy(y^2)d/dx(y)}+ye^(xy)d/dx(xy)+e^(xy)dy/dx=d/dx(xy).# #:. y^2+x{(2y)dy/dx}+ye^(xy){xdy/dx+yd/dx(x)}+e^(xy)dy/dx={xdy/dx+yd/dx(x)}.# #:. y^2+2xydy/dx+ye^(xy){xdy/dx+y}+e^(xy)dy/dx=xdy/dx+y.# #;. {2xy+xye^(xy)+e^(xy)-x}dy/dx=y-y^2-y^2e^(xy).# #:. dy/dx={y(1-y-ye^(xy))}/{2xy+xye^(xy)+e^(xy)-x}.# This Answer is fairly reasonable, but, if we replace #e^(xy)# from this, we can have better Answer as shown below : #NR.=y{1-y-(xy+1-y^2x)}=y(1-y-xy-1+y^2x)=y(xy^2-xy-y)=y^2(xy-x-1).# #DR.=2xy-x+x(xy+1-y^2x)+(xy+1-y^2x)/y,# #=(2xy^2-xy+x^2y^2+xy-y^3x^2+xy+1-y^2x)/y# #=(xy+xy^2+x^2y^2-x^2y^3+1)/y.# #rArr dy/dx={y^3(xy-x-1)}/(xy+xy^2+x^2y^2-x^2y^3+1).#

Elijah Abram · Answer

undefined To implicitly differentiate the equation -1=-y^2x+xy-ye^(xy):

1. Differentiate each term with respect to x.
2. Apply the product rule and chain rule where necessary.
3. Solve for dy/dx.

Differentiating each term:
1. For -1, the derivative is 0.
2. For -y^2x, the derivative is -y^2 - 2yx(dy/dx).
3. For xy, the derivative is y + x(dy/dx).
4. For -ye^(xy), the derivative is -y^2e^(xy) - y^3x(dy/dx) - ye^(xy).

Combine the terms involving dy/dx:
-2yx(dy/dx) + x(dy/dx) - y^3x(dy/dx) = -y^2 - y^2e^(xy)

Factor out dy/dx:
(dy/dx)(-2yx + x - y^3x) = -y^2 - y^2e^(xy)

Solve for dy/dx:
dy/dx = (-y^2 - y^2e^(xy)) / (-2yx + x - y^3x)