※ 引述《betray911015 (回頭太難)》之銘言：
: g(x) 1 cosx
: If f(x) =∫ ---------- dt , where g(x)=∫ (1+sint^2)dt
: 0 √(1+t^2) 0
: find f'(π/2), ans: -1
: 完全無頭緒，請求高手
g(x) 1
f(x) = ∫ ------------- dt
0 √(1 + t^2)
1
f'(x) = (g'(x))(------------------)
√(1 + (g(x))^2)
π 1
f'(---) = (g'(π/2))(--------------------)
2 √(1 + (g(π/2))^2)
cosx
g(x) = ∫ (1 + (sin(t))^2) dt
0
g'(x) = (-sin(x))(1 + (sin(cos(x)))^2)
π
g'(---) = (-1)(1 + (sin(cos(π/2)))^2)
2
= (-1)(1 + (sin(0))^2)
= (-1)(1 + 0^2)
= (-1)(1 + 0) = -1
π cos(π/2)
g(---) = ∫ (1 + (sin(t))^2) dt
2 0
0
= ∫ (1 + (sin(t))^2) dt = 0
0
π π 1
f'(---) = (g'(---))(---------------------)
2 2 √(1 + (g(π/2))^2)
1
= (-1)(--------------)
√(1 + 0^2)
1
= (-1)(---) = -1
1
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