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What is the derivative of #f(x)=(e^(1/x))/x^2# ?

Answer 1

Aurora Alvarado

By Quotient Rule,

#f'(x)={-1/x^2e^{1/x}cdot x^2-e^{1/x}cdot2x}/{(x^2)^2}#

by cancelling out #x^2# in the numerator,

#={-e^{1/x}-2xe^{1/x}}/{x^4}#

by factoring out #-e^{1/x}# from the numerator,

#=-{(1-2x)e^{1/x}}/{x^4}#

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Answer 2

Aurora Alvarado

To find the derivative of ( f(x) = \frac{e^{1/x}}{x^2} ), you can use the quotient rule and the chain rule. The derivative is given by:

[ f'(x) = \frac{d}{dx} \left( \frac{e^{1/x}}{x^2} \right) ] [ = \frac{e^{1/x} \cdot d}{dx} \left( \frac{1}{x^2} \right) - \frac{d}{dx} \left( e^{1/x} \right) \cdot \frac{1}{x^2} ]

Using the chain rule and the power rule for differentiation, we can find the derivatives of ( e^{1/x} ) and ( \frac{1}{x^2} ):

[ = \frac{e^{1/x} \cdot (-2x^{-3}) - e^{1/x} \cdot \left( -\frac{1}{x^2} \right)}{x^4} ] [ = \frac{-2e^{1/x}}{x^3} + \frac{e^{1/x}}{x^4} ] [ = e^{1/x} \left( \frac{1}{x^4} - \frac{2}{x^3} \right) ]

So, the derivative of ( f(x) = \frac{e^{1/x}}{x^2} ) is ( f'(x) = e^{1/x} \left( \frac{1}{x^4} - \frac{2}{x^3} \right) ).

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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.