根据导数的定义求下列函数的导数:⑴求函数y=x²+3在x=1处的导数⑵求函数y=1/x在x=a(a≠0)处的导数
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第一题:
y = x² + 3
y'|(x=1)
= lim(Δx→0) [f(1 + Δx) - f(1)]/Δx
= lim(Δx→0) {[(1 + Δx)² + 3] - 4}/Δx
= lim(Δx→0) {[1 + 2Δx + (Δx)² + 3] - 4}/Δx
= lim(Δx→0) [2Δx + (Δx)²]/Δx
= lim(Δx→0) (2 + Δx)
= 2
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y = 1/x
y'|(x=a)
= lim(Δx→0) [f(a + Δx) - f(a)]/Δx
= lim(Δx→0) [1/(a + Δx) - 1/a]/Δx
= lim(Δx→0) [a - (a + Δx)]/[aΔx(a + Δx)]
= lim(Δx→0) -Δx/[aΔx(a + Δx)]
= lim(Δx→0) -1/[a(a + Δx)]
= -1/[a(a + 0)]
= -1/a²
y = x² + 3
y'|(x=1)
= lim(Δx→0) [f(1 + Δx) - f(1)]/Δx
= lim(Δx→0) {[(1 + Δx)² + 3] - 4}/Δx
= lim(Δx→0) {[1 + 2Δx + (Δx)² + 3] - 4}/Δx
= lim(Δx→0) [2Δx + (Δx)²]/Δx
= lim(Δx→0) (2 + Δx)
= 2
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y = 1/x
y'|(x=a)
= lim(Δx→0) [f(a + Δx) - f(a)]/Δx
= lim(Δx→0) [1/(a + Δx) - 1/a]/Δx
= lim(Δx→0) [a - (a + Δx)]/[aΔx(a + Δx)]
= lim(Δx→0) -Δx/[aΔx(a + Δx)]
= lim(Δx→0) -1/[a(a + Δx)]
= -1/[a(a + 0)]
= -1/a²
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