Mathematics
Senior High
Resolved
(2)について
微分法①
黄色いマーカー部分がどうしてこうなるのか分からないです。
よく出る
(1) 次の値を求めよ。
h²-5h
lim
h→0 h
SIS) 02
(2) 次の関数の, 与えられた値における微分係数を求めよ。
f(x)=x2-2x-5, x=-1
2
h²-5h
h
TA
(1) lim-
h→0
(2) f'(-1)= lim
h→0
=
=
= lim (h-5) = -5
h→0
=
(3) 2=(4,
- lim
h→0
lim-
h→0
= = lim
f-1+h)-f(-1)
h
h²-4h
h
h→0
{(−1+h)²-2(−1+h)−5}-{(−1)²−2∙(−1)−5}
h
(1-2h+h²+2-2h-5)-(-2)
h
【】 f'(-1)= lim
b→-1
= lim
6→-1
= lim
b→-1
= lim
=
b→-1
= lim
6--1
......
h→0
A
lim(h-4)=-4
• (*)
f(b) f(-1)
b-(-1)
6²-26-3
b+1
TURNSEACRHoain
62-26-5-(-2)
b+1
(b+1)(b-3)
b+1
A
(b²-2b-5)-{(−1)²−2∙(−1)−5} −A£+22+²
b+1
·()
A
曲
る
28=(+18) mil ka
$+18=(3+√x} mil=
-= lim (b-3)=-4 ·()
b-1
Answers
.jpg?Expires=1776057483&Signature=xbuGtfJURKj188tSEP7tT8vd7sEcCtOayglfedL-goXv9jvd23~hMpkqNHPHeMAOFHLf8TvCPF-jSfoADz3xf2h33P~gDK~aF3Qj7tCMbwoTQefQSQMVAnbEcBGlLUssCBN-ma0Ot44KYgwPCl44WD6PaAzFGbaY9LFvuSHXEgB3F2BK5yEf~0WOQy-hnvtqxDV4tHRRR81s1RAqHiiOAuoryBh6L3Slaydq2iHlCtqybPhKfr6FsnrysSq01FgkCu5I1ayX7-qVcjrZGdI0U-N2eRYuY5eeV64uMFiP9CJoZ5NXhr4clSJmUSUg4MuM-YcomDdv6G2Suei1j3-L4Q__&Key-Pair-Id=K2W722D70GJS8W)
Were you able to resolve your confusion?
Users viewing this question
are also looking at these questions 😉
ありがとうございます！！理解しました！