Practice problem
Find the derivative of g(t)=tanh(t2+1)g(t)=\tanh(\sqrt{t^2+1})g(t)=tanh(t2+1)
g′(t)=−1tt2+1g'(t)=-\frac{1}{t\sqrt{t^2+1}}g′(t)=−tt2+11
g′(t)=t sech2(t2+1)g'(t)=t\ \text{sech}^2(\sqrt{t^2+1})g′(t)=t sech2(t2+1)
g′(t)=sech2(t2+1)t2+1g'(t)=\frac{\text{sech}^2(\sqrt{t^2+1})}{t^2+1}g′(t)=t2+1sech2(t2+1)
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”.
