# What is the second derivative of #x/(2+ e^x)#?

So,

Aggregating them in a multiplication as stated by chain rule:

Distributing the factors, we get the following, which allows us to operate some cancelling, as I will remark in collors:

Final simplification:

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To find the second derivative of ( \frac{x}{2+ e^x} ), first differentiate it once using the quotient rule, then differentiate the result again using the quotient rule and chain rule.

First derivative: [ \frac{d}{dx} \left( \frac{x}{2+ e^x} \right) = \frac{(2+e^x)(1) - (x)(e^x)}{(2+e^x)^2} ]

Second derivative: [ \frac{d^2}{dx^2} \left( \frac{x}{2+ e^x} \right) = \frac{(2+e^x)(0) - (1)(e^x) - (x)(e^x)}{(2+e^x)^2} - \frac{(2+e^x)^2(e^x) - 2(2+e^x)(1)(e^x)}{(2+e^x)^4} ]

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To find the second derivative of ( \frac{x}{2 + e^x} ), you first need to find the first derivative, then differentiate it again.

( f(x) = \frac{x}{2 + e^x} )

First derivative, ( f'(x) = \frac{(2 + e^x)(1) - (x)(0 + e^x)}{(2 + e^x)^2} )

( f'(x) = \frac{2 + e^x - xe^x}{(2 + e^x)^2} )

Now, find the second derivative by differentiating ( f'(x) ).

( f''(x) = \frac{((2 + e^x)^2)(0 + e^x) - (2 + e^x - xe^x)(2e^x)}{(2 + e^x)^4} )

( f''(x) = \frac{(2 + e^x)(e^x) - (2e^x + e^{2x} - 2xe^x)}{(2 + e^x)^3} )

( f''(x) = \frac{2e^x + e^{2x} - 2xe^x - 2e^x}{(2 + e^x)^3} )

( f''(x) = \frac{e^x + e^{2x} - 2xe^x}{(2 + e^x)^3} )

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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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