a)$$\eqalign{ f(x) = {2x^{-3/2}} \implies {dy \over dx} = -3x^{-5/2} &= -{3 \over \sqrt{x^5}} }$$ b) $$\eqalign{ x = 2t + 1 \implies {dx \over dt} = 2 \\ y = t^2 \implies {dy \over dt} = 2t \\ {dy \over dx} = {dy \over dt} \div {dx \over dt} \\ \implies 2t \div 2 = t }$$