How do you differentiate #y=e^x/(1+x)#?

Answer 1

The answer is #=(xe^x)/(1+x)^2#

We need the differentiation of a quotient

#(u/v)'=(u'v-uv')/v^2#
Here, #u=e^x#, #=>#, #u'=e^x#
#v=1+x#, #=>#, #v'=1#

Therefore,

#dy/dx=(e^x(x+1)-e^x)/(1+x)^2#
#=(xe^x+e^x-e^x)/(1+x)^2#
#=(xe^x)/(1+x)^2#
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Answer 2

To differentiate the function ( y = \frac{e^x}{1+x} ), you can use the quotient rule, which states that if you have a function of the form ( \frac{u(x)}{v(x)} ), then its derivative is given by:

[ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - v'(x)u(x)}{[v(x)]^2} ]

In this case, ( u(x) = e^x ) and ( v(x) = 1 + x ). Now, find the derivatives of ( u(x) ) and ( v(x) ), denoted as ( u'(x) ) and ( v'(x) ), respectively, and substitute them into the quotient rule formula.[ y' = \frac{(e^x)(1+x) - (e^x)(1)}{(1+x)^2} = \frac{e^x(1+x) - e^x}{(1+x)^2} ] [ y' = \frac{e^x + xe^x - e^x}{(1+x)^2} = \frac{xe^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.

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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