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How do you implicitly differentiate # e^(3x)/y^2=xsin(x-2y) #?

Answer 1

Charles Anctil

See below.

#frac{D_x e^(3x) y^2 - e^(3x) D_x y^2}{y^4} = 1sin(x - 2y) + x cos (x - 2y) (1 - 2y')#

#frac{3 e^(3x) y^2 - e^(3x) 2y'y}{y^4} = sin(x - 2y) + x cos (x - 2y) (1 - 2y')#

# y' = frac{y^4 sin(x - 2y) + xy^4 cos (x - 2y) - 3 e^(3x) y^2}{2xy^4 cos (x - 2y) - e^(3x) 2y}#

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Answer 2

Christopher Agresta

To implicitly differentiate ( \frac{e^{3x}}{y^2} = x \sin(x - 2y) ):

Differentiate both sides with respect to ( x ).
Use the chain rule, product rule, and quotient rule as needed.

The result after differentiation will be:

[ \frac{3e^{3x}}{y^2} - \frac{2e^{3x}}{y^3} \frac{dy}{dx} = \sin(x - 2y) + x \cos(x - 2y) \frac{dx}{dx} - x \sin(x - 2y) \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.