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How do you find the derivative of #y=e^x/x#?

Answer 1

Charles Arocha

You can use the Quotinet Rule where: if #y=f(x)/g(x) -> y'=[f'(x)g(x)-f(x)g'(x)]/[g(x)]^2#

So you get: #y'=(xe^x-1e^x)/x^2=# #=e^x(x-1)/x^2#

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Answer 2

Christopher Agresta

To find the derivative of ( y = \frac{e^x}{x} ), you can use the quotient rule. The quotient rule states that for two functions ( u ) and ( v ) where ( y = \frac{u}{v} ), the derivative ( \frac{dy}{dx} ) is given by:

[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} ]

In this case, ( u = e^x ) and ( v = x ). The derivatives of ( u ) and ( v ) with respect to ( x ) are ( \frac{du}{dx} = e^x ) and ( \frac{dv}{dx} = 1 ), respectively. Plugging these into the quotient rule formula gives:

[ \frac{dy}{dx} = \frac{x \cdot e^x - e^x \cdot 1}{x^2} ] [ \frac{dy}{dx} = \frac{x \cdot e^x - e^x}{x^2} ] [ \frac{dy}{dx} = \frac{e^x(x - 1)}{x^2} ]

So, the derivative of ( y = \frac{e^x}{x} ) is ( \frac{e^x(x - 1)}{x^2} ).

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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.