How do you calculate the derivative for #e^x/(e^x+1)(x+1)#?

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Answer 1 · 2018-05-10T16:58:07

#f'(x)=1-1/((e^x+1)²)#

#f(x)=e^x/(e^x+1)(x+1)#

#f'(x)=(e^x/(e^x+1))^'(x+1)+e^x/(e^x+1)#

Now : #(u/v)^'=(u'v-uv')/(v²)#

Here :#f'(x)=(e^x(e^x+1)-e^(2x))/((e^x+1)²)+e^x/(e^x+1)#

#f'(x)=e^x/((e^x+1)²)+e^x/(e^x+1)#

#f'(x)=e^x/((e^x+1)²)+(e^(2x)+e^x)/((e^x+1)²)#

#f'(x)=(e^(2x)+2e^x)/((e^x+1)²)#

#f'(x)=(cancel((e^x+1)²)-1)/cancel((e^x+1)²)#

#f'(x)=1-1/((e^x+1)²)#

\0/ here's our answer!