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

Answer 1

#dy/dx=(e^(-x))/(1+e^(-x))^2,(d^2y)/(dx^2)=(e^(-2x)-e^(-x))/(1+e^(-x))^3#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(e^(-x))=-e^(-x))color(white)(2/2)|)))#

There are 2 approaches to differentiating this function.

#(1)" Using the quotient rule"#
#(2)" expressing " y=(1+e^(-x))^-1" and use chain rule"#

I'll use approach (1) you could perhaps try approach (2). The result will be the same.

differentiate using the #color(blue)"quotient rule"#
#" Given " y=(g(x))/(h(x))" then"#
#color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(2/2)|)))#
#g(x)=1rArrg'(x)=0#
#h(x)=1+e^(-x)rArrh'(x)=-e^(-x)# #color(blue)"------------------------------------------------------------"#
#rArrdy/dx=((1+e^-x).0-1.(-e^-x))/(1+e^-x)^2=(e^-x)/(1+e^-x)^2#
#" To find " (d^2y)/(dx^2)" differentiate " dy/dx#
differentiate using the #color(blue)"quotient rule/chain rule"#
#"here " g(x)=e^-xrArrg'(x)=-e^-x#
#h(x)=(1+e^-x)^2rArrh'(x)=2(1+e^-x).(-e^-x)rarr#
#rArrh'(x)=-2e^-x(1+e^-x)# #color(blue)"---------------------------------------------------------------"#
#(d^2y)/(dx^2)=((1+e^-x)^2(-e^-x)-(e^-x).(-2e^-x(1+e^-x)))/(1+e^-x)^4#
#=(-e^-x(1+e^-x)^2+2e^(-2x)(1+e^-x))/(1+e^-x)^4#
#=(e^-x(1+e^-x)(2e^-x-1-e^-x))/(1+e^-x)^4larr" factoring"#
#=(e^-xcancel((1+e^-x))(e^-x-1))/(cancel((1+e^-x)^3#
#=(e^(-2x)-e^-x)/(1+e^-x)^3#
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Answer 2

To find the first derivative of ( y = \frac{1}{1 + e^{-x}} ), we use the chain rule.

Let ( u = 1 + e^{-x} ), then ( y = u^{-1} ).

Thus, using the chain rule, the derivative of ( y ) with respect to ( x ) is:

[ \frac{dy}{dx} = \frac{d}{dx} \left( u^{-1} \right) = -u^{-2} \frac{du}{dx} ]

Now, differentiate ( u ) with respect to ( x ):

[ \frac{du}{dx} = \frac{d}{dx} (1 + e^{-x}) = 0 - e^{-x}(-1) = e^{-x} ]

Substitute ( u ) and ( \frac{du}{dx} ) back into the derivative:

[ \frac{dy}{dx} = -\frac{1}{(1 + e^{-x})^2} \cdot e^{-x} ]

To find the second derivative, differentiate ( \frac{dy}{dx} ) with respect to ( x ):

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

Using the product rule and chain rule, the derivative is:

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

Now, differentiate each term:

[ \frac{d}{dx} \left( \frac{1}{(1 + e^{-x})^2} \right) = \frac{2}{(1 + e^{-x})^3} \cdot e^{-x} ]

[ \frac{d}{dx}(e^{-x}) = -e^{-x} ]

Substitute these derivatives back:

[ \frac{d^2y}{dx^2} = -\frac{2}{(1 + e^{-x})^3} \cdot e^{-2x} + \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.

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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