數學 - derivative

By chain rule,
i.e. d/dx [ p(x) q(x)] = q(x) d/dx [p(x)] + p(x) d/dx [q(x)]

1/
d/dt [ (t^2+1)sint + ln(e^t cost) ]
= d/dt [ (t^2+1)sint ] + d/dt[ ln(e^t cost) ]
= [ (t^2+1) d/dt (sint) + sint d/dt (t^2+1) ] + [1/(e^t cost) * d/dt(e^t cost)]
= [(t^2+1) cost + 2t sint ] + [1/(e^t cost) * e^t cost * d/dt (t cost)]
= [(t^2+1) cost + 2t sint ] + [t * d/dt(cos t) + cost * d/dt (t)]
= [(t^2+1) cost + 2t sint ] + ( -t sint + cost)
= (t^2+2) cost + t sint

2/
d/dp [ (1+psinp) / (e^p + p)]
= [(e^p + p) d/dp (1+psinp) - (1+psinp) d/dp(e^p + p)] / (e^p + p)^2
= {(e^p + p) [p * d/dp (sinp) + sinp d/dp (p)] - (1+psinp) [e^p d/dp(p) + d/dp(p)]} / (e^p + p)^2
= {(e^p + p) (p cosp) + sinp - (1+psinp) (e^p + 1)}/ (e^p + p)^2


3/
d/dt [sint + te^(t+lnt)]
= d/dt [sint ]+d/dt [ te^(t+lnt)]
= cost + e^(t+lnt) d/dt (t) + t d/dt [ e^(t+lnt)]
= cost + e^(t+lnt) + t [ e^(t+lnt)] d/dt [ (t+lnt)]
= cost + e^(t+lnt) + t [ e^(t+lnt) (1+ 1/t)]
= cost + e^(t+lnt) + e^(t+lnt) (t + 1)
= cost + (t + 2) e^(t+lnt)

2007-10-18 14:29:43 補充：
last row for Q2 should be = {(e^p p) (p cosp sinp) - (1 psinp) (e^p 1)}/ (e^p p)^2

Have you learnt the following formulae?

d/dx [ p(x) q(x)] = q(x) dp(x)/dx + p(x) dq(x)/dx

This should help.