Is f bijective and if so, calculate the n order derivative of it\'s inverse; f(x)=[(e^x)-(e^- x)]/[(e^x)+(e^-x)]. Solution We\'ll re-write f(x)=[e^x-(1/e^x)]/[e^x+(1/e^x)] f(x)=(e^2x-1)/(e^2x+1)=1 - 2/(e^2x+1) 2/(e^2x+1)<1, f(x)>0, so f(x) is strictly increasing, so f is bijective. f(x)=y y=1 - 2/(e^2x+1) 1-y=2/(e^2x+1) ln(1-y)=ln[2/(e^2x+1)] ln(1-y)=ln2-ln(e^2x+1) ln(1-y)-ln2=ln(e^2x+1) ln[(1-y)/2]=ln(e^2x+1) (1-y)/2=e^2x+1 (1-y)/2-1=e^2x ln[(1-y)/2-1]=2xlne {ln[(1-y)/2-1]}/2=x.

1. Is f bijective and if so, calculate the n order derivative of it's inverse; f(x)=[(e^x)-(e^-
x)]/[(e^x)+(e^-x)].
Solution
We'll re-write f(x)=[e^x-(1/e^x)]/[e^x+(1/e^x)]
f(x)=(e^2x-1)/(e^2x+1)=1 - 2/(e^2x+1)
2/(e^2x+1)<1, f(x)>0, so f(x) is strictly increasing, so f is bijective.
f(x)=y
y=1 - 2/(e^2x+1)
1-y=2/(e^2x+1)
ln(1-y)=ln[2/(e^2x+1)]
ln(1-y)=ln2-ln(e^2x+1)
ln(1-y)-ln2=ln(e^2x+1)
ln[(1-y)/2]=ln(e^2x+1)
(1-y)/2=e^2x+1
(1-y)/2-1=e^2x
ln[(1-y)/2-1]=2xlne
{ln[(1-y)/2-1]}/2=x