How do you calculate the derivative for #e^x/(e^x+1)(x+1)#?
Using product rule :
\0/ here's our answer!
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To calculate the derivative of the given function, ( \frac{e^x}{(e^x+1)(x+1)} ), you can use the quotient rule of differentiation. The quotient rule states that if you have a function of the form ( \frac{f(x)}{g(x)} ), the derivative is given by:
[ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} ]
For our function, ( f(x) = e^x ) and ( g(x) = (e^x+1)(x+1) ). Taking the derivatives of ( f(x) ) and ( g(x) ), we get:
[ f'(x) = e^x ] [ g'(x) = (e^x)(x+1) + (e^x+1)(1) ]
Now, we can plug these into the quotient rule formula:
[ \frac{d}{dx}\left(\frac{e^x}{(e^x+1)(x+1)}\right) = \frac{(e^x)((e^x+1)(x+1)) - (e^x)(e^x) - (e^x)}{((e^x+1)(x+1))^2} ]
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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.
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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