How do you differentiate $f(x)=8e^(x^2)/(e^x+1)$ using the chain rule?

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Answer 1 · 2016-02-13T18:25:10

The only trick here is that $(e^(x^2))'=e^(x^2)*(x^2)'=e^(x^2)*2x$
Final derivative is:

$f'(x)=8e^(x^2)(2x*(e^x+1)-e^x)/(e^x+1)^2$
or
$f'(x)=8e^(x^2)(e^x*(2x-1)+2x+1)/(e^x+1)^2$

$f(x)=8(e^(x^2))/(e^x+1)$

$f'(x)=8((e^(x^2))'(e^x+1)-e^(x^2)(e^x+1)')/(e^x+1)^2$

$f'(x)=8(e^(x^2)*(x^2)'(e^x+1)-e^(x^2)*e^x)/(e^x+1)^2$

$f'(x)=8(e^(x^2)2x*(e^x+1)-e^(x^2)*e^x)/(e^x+1)^2$

$f'(x)=8(e^(x^2)(2x*(e^x+1)-e^x))/(e^x+1)^2$

$f'(x)=8e^(x^2)(2x*(e^x+1)-e^x)/(e^x+1)^2$

or (if you want to factor $e^x$ in the nominator)

$f'(x)=8e^(x^2)(e^x*(2x-1)+2x+1)/(e^x+1)^2$

Note: if you want to study the sign, you are gonna have a bad time. Just look at the graph:

graph{8(e^(x^2))/(e^x+1) [-50.25, 53.75, -2.3, 49.76]}

How do you differentiate f(x)=8e^(x^2)/(e^x+1)f(x)=8e^(x^2)/(e^x+1) using the chain rule?