# What is the derivative of #y=(x/e^x)LnX#?

An equation of this form is deived like this:

An equation of this form is derived like this:

Knowing that:

Therefore:

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To find the derivative of ( y = \frac{x}{e^x} \ln(x) ):

Use the product rule:

[ (uv)' = u'v + uv' ]

Where: [ u = \frac{x}{e^x} ] [ v = \ln(x) ]

[ u' = \frac{d}{dx} \left( \frac{x}{e^x} \right) ] [ v' = \frac{d}{dx} \ln(x) ]

Using quotient rule for ( u' ):

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

[ \frac{d}{dx} \ln(x) = \frac{1}{x} ]

Now, apply the product rule:

[ y' = \left( \frac{e^x - x e^x}{e^{2x}} \right) \ln(x) + \frac{x}{e^x} \cdot \frac{1}{x} ]

Simplify:

[ y' = \frac{e^x - x e^x}{e^{2x}} \ln(x) + \frac{1}{e^x} ]

That's the derivative of ( y = \frac{x}{e^x} \ln(x) ).

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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.

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.

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.

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