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What is the implicit derivative of #1= xy-e^(xy) #?

Answer 1

Aurora Alvarado

#y' = - y/x, \qquad x,y \ne 0#

# 0 = d/dx (xy) - d/dx e^{xy}#

first term, using product rule:

# d/dx (xy) = y + x y'#

second term using chain rule:

#d/dx e^{xy} = d/dx(xy) * e^{xy}#

with product rule this is #d/dx(xy) * e^{xy} = (y + x y') * e^{xy}#

combining terms

#d/dx (xy) - d/dx e^{xy} = (y + x y') - (y + x y') * e^{xy} #

#0 = (y + x y') ( 1 - e^{xy})#

ignoring trivial solution #( 1 - e^{xy}) = 0 \implies x = 0, or y = 0# leaves

# y + x y' = 0 \implies y' = - y/x, \qquad x,y \ne 0#

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Answer 2

Emily Ammerman

To find the implicit derivative of ( 1 = xy - e^{xy} ), we differentiate both sides of the equation with respect to ( x ) using implicit differentiation.

( \frac{d}{dx}(1) = \frac{d}{dx}(xy - e^{xy}) )

( 0 = y + x\frac{dy}{dx} - e^{xy}(y + x\frac{dy}{dx}) )

( 0 = y + x\frac{dy}{dx} - ye^{xy} - x(e^{xy})\frac{dy}{dx} )

( 0 = y + x\frac{dy}{dx} - ye^{xy} - xe^{xy}\frac{dy}{dx} )

Now, we solve for ( \frac{dy}{dx} ):

( x\frac{dy}{dx} - xe^{xy}\frac{dy}{dx} = ye^{xy} - y )

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

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

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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.