# How do you find the derivative of #x/(e^(2x))#?

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we can arrange this function so that we can use the product rule

the product rule

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To find the derivative of ( \frac{x}{e^{2x}} ), you can use the quotient rule. The quotient rule states that if you have a function of the form ( \frac{f(x)}{g(x)} ), then its derivative is given by ( \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} ).

In this case: ( f(x) = x ) ( g(x) = e^{2x} )

Now, find the derivatives of ( f(x) ) and ( g(x) ): ( f'(x) = 1 ) ( g'(x) = 2e^{2x} )

Now, apply the quotient rule: ( \frac{d}{dx}\left(\frac{x}{e^{2x}}\right) = \frac{(1)(e^{2x}) - (x)(2e^{2x})}{(e^{2x})^2} )

Simplify: ( \frac{e^{2x} - 2xe^{2x}}{e^{4x}} )

Thus, the derivative of ( \frac{x}{e^{2x}} ) is ( \frac{e^{2x} - 2xe^{2x}}{e^{4x}} ).

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To find the derivative of ( \frac{x}{e^{2x}} ), you can use the quotient rule, which states that if you have a function ( \frac{f(x)}{g(x)} ), the derivative is given by:

[ \left( \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{(g(x))^2} \right) ]

Here, ( f(x) = x ) and ( g(x) = e^{2x} ).

Now, find the derivatives of ( f(x) ) and ( g(x) ), which are ( f'(x) ) and ( g'(x) ), respectively.

[ f'(x) = 1 ] [ g'(x) = 2e^{2x} ]

Now, substitute these derivatives into the quotient rule formula:

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

[ = \frac{e^{2x} - 2xe^{2x}}{e^{4x}} ]

So, the derivative of ( \frac{x}{e^{2x}} ) is ( \frac{e^{2x} - 2xe^{2x}}{e^{4x}} ).

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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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