Practice problem
Find the derivative of g(t)=tanh−1(t2+1)g(t)=\tanh^{-1}(\sqrt{t^2+1})g(t)=tanh−1(t2+1)
g′(t)=−tt2+1g'(t)=-\frac{t}{t^2+1}g′(t)=−t2+1t
g′(t)=−1tt2+1g'(t)=-\frac{1}{t\sqrt{t^2+1}}g′(t)=−tt2+11
g′(t)=1−tt2+1g'(t)=\frac{1-t}{\sqrt{t^2+1}}g′(t)=t2+11−t
g′(t)=t sech2(t2+1)t2+1g'(t)=\frac{t\ \text{sech}^2(\sqrt{t^2+1})}{\sqrt{t^2+1}}g′(t)=t2+1t sech2(t2+1)
Type a concept, like “slope” or “photosynthesis”.
