We will first find f ` ( x ) using the chain rule:
f ` ( x ) = [tex] \frac{1}{2 \sqrt{e ^{2x}+1 } } * 2e^{2x} = \frac{e^{2x} }{ \sqrt{e ^{2x} +1} } [/tex]
f ` ( 0 ) = [tex] \frac{e ^{0} }{ \sqrt{e^{0}+1 } } = \frac{1}{ \sqrt{2} } [/tex]
Answer: f ` ( 0 ) = 1/√2 ( or: √2/2 )
