How do you implicitly differentiate #-y^2=e^(2x-4y)-2y/x #?

Answer 1

#y'=(x^2*e^(2x-4y)+y)/(2x^2*e^(2x-4y) +x-x^2y)#

Start differentiating with respect to x both sides of the equation #-y^2=e^(2x-4y) -2*y/x#
#d/dx(-y^2)=d/dx(e^(2x-4y)) -d/dx(2*y/x)#
#(-2y*y')=(e^(2x-4y)*d/dx(2x-4y)) -(2*d/dx(y/x))#
#-2y*y'=e^(2x-4y)(2-4y') -2*((xy'-y*1)/x^2)#

Simplify by dividing by 2

#-y*y'=e^(2x-4y)(1-2y') -((xy'-y*1)/x^2)#
#-y*y'=e^(2x-4y)-2*e^(2x-4y)y' -(xy')/x^2+(y)/x^2#
#-y*y'=e^(2x-4y)-2*e^(2x-4y)y' -(y')/x+y/x^2#
#2*e^(2x-4y)y' +(y')/x-y*y'=e^(2x-4y)+y/x^2#

Factor the y'

#(2*e^(2x-4y) +1/x-y)y'=e^(2x-4y)+y/x^2#
Divide both sides by #(2*e^(2x-4y) +1/x-y)#
#((2*e^(2x-4y) +1/x-y)y')/((2*e^(2x-4y) +1/x-y))=(e^(2x-4y)+y/x^2)/(2*e^(2x-4y) +1/x-y)#
#cancel((2*e^(2x-4y) +1/x-y)y')/cancel((2*e^(2x-4y) +1/x-y))=(e^(2x-4y)+y/x^2)/(2*e^(2x-4y) +1/x-y)#
#y'=(e^(2x-4y)+y/x^2)/(2*e^(2x-4y) +1/x-y)#
Simplify by multiplying the right side by #x^2/x^2#
#y'=((e^(2x-4y)+y/x^2))/((2*e^(2x-4y) +1/x-y))*x^2/x^2#
#y'=(x^2*e^(2x-4y)+y)/(2x^2*e^(2x-4y) +x-x^2y)#

God bless....I hope the explanation is useful.

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Answer 2

To implicitly differentiate -y^2 = e^(2x - 4y) - (2y/x) with respect to x:

  1. Differentiate both sides of the equation with respect to x.
  2. Apply the chain rule and the product rule where necessary.
  3. Solve for dy/dx.

The result is: dy/dx = (4y^2 - 2xe^(2x - 4y) + 2e^(2x - 4y)) / (4y + 2x^2/y)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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