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

Question

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

Paisley Johnson · Accepted Answer

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

Take the derivative of the equation except for every time you differentiate #y#, it becomes #dy/dx#. You will also have to use the product rule.

Note: #e^(x-2y)=(e^x)(e^-(2y))#. So the equation is actually:

#y=xy^2+x^2(e^x)(e^-(2y))# so you will have to use the product rule twice for the second term:

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

Use the product rule again:

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

Distribute the x^2:

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

Separate the terms with #dy/dx#

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

Factor out the #dy/dx#:

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

Divide both sides by #(1-2xy+2x^2e^(x-2y))#:

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

Isaiah Allen · Answer

#d/dx( y= xy^2 + x  ^2 e^(x -2y))# # implies y' = y^2 + 2 x y y' + 2 x  e^(x -2y) + x^2 (color(red)(1)- 2y') e^(x -2y)# # implies y'(1-  2 x y + 2x^2  e^(x -2y) ) = y^2  + (color(red)(x^2 +) 2 x)  e^(x -2y) # # implies y'= ( y^2  +  (color(red)(x^2 +) 2 x)   e^(x -2y) )/(1-  2 x y + 2x^2  e^(x -2y) )  #

Isaiah Allen · Answer

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

Using the Implicit Function Theorem we have: # dy/dx = -((partial f)/dx)/((partial f)/dy) # Where: # f(x,y) = 0 # We have: # \ \ \ \ \ y=xy^2 + x^2e^(x-2y) # # :. y=xy^2 + x^2e^x e^(-2y) # Let: # f(x,y)= xy^2 + x^2e^x e^(-2y)-y # Then the partial derivatives are: # (partial f)/(partial x) = y^2+(x^2)(e^x e^(-2y))+(2x)(e^x e^(-2y)) # # \ \ \ \ \ \ \ = y^2+(x^2)(e^(x-2y))+(2x)(e^(x-2y)) # # \ \ \ \ \ \ \ = y^2+(x^2+2x)e^(x-2y) # And: # (partial f)/(partial y) = 2xy-2x^2e^x e^(-2y)-1 # # \ \ \ \ \ \ \ = 2xy-2x^2e^(x-2y)-1 # And so: # dy/dx = -(y^2+(x^2+2x)e^(x-2y))/(2xy-2x^2e^(x-2y)-1)# # \ \ \ \ \ \ = (y^2+(x^2+2x)e^(x-2y))/(1+2x^2e^(x-2y)-2xy)#

Samantha Adkison · Answer

undefined To implicitly differentiate $ y = xy^2 + x^2e^{x - 2y} $:

1. Differentiate each term with respect to $ x $.
2. Apply the product rule for the term $ xy^2 $ and the chain rule for the term $ x^2e^{x - 2y} $.
3. Simplify the resulting expression.

The implicit differentiation of the given function yields:

$$ \frac{dy}{dx} = y^2 + 2xy\frac{dy}{dx} + 2xe^{x - 2y} - 2x\frac{dy}{dx}e^{x - 2y} $$

Now, solve for $ \frac{dy}{dx} $:

$$ \frac{dy}{dx} - 2xy\frac{dy}{dx} + 2x\frac{dy}{dx}e^{x - 2y} = y^2 + 2xe^{x - 2y} $$

$$ \frac{dy}{dx}(1 - 2xy + 2xe^{x - 2y}) = y^2 + 2xe^{x - 2y} $$

$$ \frac{dy}{dx} = \frac{y^2 + 2xe^{x - 2y}}{1 - 2xy + 2xe^{x - 2y}} $$