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How do you find the derivative of #y=1/(1+e^x)#?

Answer 1

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

Differentiate using the chain rule Given #y=f(g(x))# #dy/dx=f'(g(x))*(g'(x)) larr#chain rule #y=(1+e^x)^-1# #dy/dx=(-(1+e^x)^-2)*e^x# #=-e^x/(1+e^x)^2#

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Answer 2

Charles Arocha

To find the derivative of ( y = \frac{1}{1 + e^x} ), you can use the quotient rule. The quotient rule states that if you have a function in the form ( \frac{u(x)}{v(x)} ), then its derivative is given by ( \frac{u'v - uv'}{(v)^2} ). Applying this rule to the given function:

( u(x) = 1 ) ( v(x) = 1 + e^x )

( u'(x) = 0 ) ( v'(x) = e^x )

Now plug these values into the quotient rule formula:

( y' = \frac{(0)(1 + e^x) - (1)(e^x)}{(1 + e^x)^2} )

Simplify:

( y' = \frac{-e^x}{(1 + e^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.