Find y′ if:
y=tan−1(4x) y=sin(tan−1x)
Solution (a) Let y=tan−1u and u=4x. Then dydu=11+u2 and dudx=4. By the Chain Rule, y′=dydx=dydu⋅dudx=11+u2⋅4=41+16x2
(b) Let y=sinu and u=tan−1x. Then dydu=cosu and dudx=11+x2. By the Chain Rule, y′=dydx=dydu⋅dudx=cosu⋅11+x2=cos(tan−1x)1+x2