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

Answer 1

Ezekiel Alexander

#(dy)/(dx)=(4-y)/[4-2y+x+2e^(2y#

Firstly multiply everything by #" "(y-x)" "# to remove the award denominator.

assuming #y!=x#

#4(y-x)=y(y-x)-e^(2y)#

#4y-4x=y^2-xy-e^(2y)#

now differentiate.

#d/(dx)(4y-4x)=d/(dx)(y^2-xy-e^(2y))#

#4(dy)/(dx)-4=2y(dy)/(dx)-(y+x(dy)/(dx))-2(dy)/(dx)e^(2y)#

now rearrange.

#4(dy)/(dx)-2y(dy)/(dx)+x(dy)/(dx)+2(dy)/(dx)e^(2y)=4-y#

#(dy)/(dx)[4-2y+x+2e^(2y)]=4-y#

#(dy)/(dx)=(4-y)/[4-2y+x+2e^(2y#

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Answer 2

Charles Anctil

To implicitly differentiate (4 = y - \frac{e^{2y}}{y-x}), first differentiate both sides of the equation with respect to (x).

The derivative of (4) with respect to (x) is (0) since it's a constant.

For the left side, differentiate (y) with respect to (x) to get (\frac{dy}{dx}).

For the right side, you need to use the quotient rule. Let (u = y - x) and (v = e^{2y}). Then apply the quotient rule (\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}).

After finding the derivatives, rearrange the equation to solve for (\frac{dy}{dx}).

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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.