Practice problem
Find the derivative. g(t)=cot−1(2t)+cot−1(2t)g(t)=\cot^{-1}(2t)+\cot^{-1}\bigg(\frac2t\bigg)g(t)=cot−1(2t)+cot−1(t2)
g′(t)=6−6t24t4+17t2+4g'(t)=\frac{6-6t^2}{4t^4+17t^2+4}g′(t)=4t4+17t2+46−6t2
g′(t)=3t2−34t4−8t2+1g'(t)=\frac{3t^2-3}{4t^4-8t^2+1}g′(t)=4t4−8t2+13t2−3
g′(t)=2+2t21−t4−5g'(t)=\frac{\sqrt{2+2t^2}}{1-\sqrt{t^4-5}}g′(t)=1−t4−52+2t2
g′(t)=2g'(t)=2g′(t)=2
Type a concept, like “slope” or “photosynthesis”.
