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

Answer 1

Evelyn Agard

#e^x/(e^x+1#

differentiate using the #color(blue)"chain rule"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(a/a)|)))........ (A)#

let #color(blue)(u=e^x+1)rArr(du)/(dx)=e^x#

and #y=lncolor(blue)(u)rArr(dy)/(du)=1/color(blue)(u)#

substitute these values into (A) and convert #color(blue)(u)# back into x

#rArrdy/dx=1/color(blue)(u).e^x=e^x/(e^x+1)#

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Answer 2

Charles Anctil

To find the derivative of ( \ln(e^x + 1) ), you can use the chain rule. The derivative is:

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

Now, differentiate ( e^x + 1 ) with respect to ( x ):

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

Substitute this result back into the expression:

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

Simplify:

[ \frac{d}{dx} \left( \ln(e^x + 1) \right) = \frac{e^x}{e^x + 1} ]

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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.