pure math 問題,數學勁人請進

g(t) = t1/5 [sin(ln t1/2)]3

Differentiate both sides w.r.t. t

g’(t) = [sin(ln t1/2)]3 dt1/5/dt + t1/5 d/dt [sin(ln t1/2)]3

­­By chain rule,

g’(t) = [sin(ln t1/2)]3 1/(5t4/5) + t1/5 3[sin(ln t1/2)]2‧cos(ln t1/2)‧1/t1/2‧1/(2t1/2)

g’(t) = [sin(ln t1/2)]3 / (5t4/5) + 3[sin(ln t1/2)]2‧cos(ln t1/2) / 2t(1/2 + 1/2 – 1/5)

g’(t) = [sin(ln t1/2)]3 / (5t4/5) + 3[sin(ln t1/2)]2‧cos(ln t1/2) / (2t4/5)

g’(t) = [2sin3(ln t1/2) + 15sin2(ln t1/2)cos(ln t1/2)] / (10t4/5)