用导数定义求x^a[sin(1/x)]的导数

Question

用导数定义求x^a[sin(1/x)]的导数

Answer 1

楼上用的是链式法则,不是定义.
f(x) = x^a[sin(1/x)] ,
用定义求应该是 f'(x) = [ f(x+Δx) - f(x) ] / Δx = [ (x+Δx)^a[sin(1/(x+Δx))] - x^a[sin(1/x)] ] / Δx
= { (x+Δx) ^ a[sin(1/x)] ^ [sin(1/(x+Δx) / sin(1/x) ] - x^a[sin(1/x)] } / Δx
∵ x+Δx = x * ((x+Δx)/x) = x * (1+ Δx /x )
= { x^a[sin(1/x)] ^[sin(1/(x+Δx) / sin(1/x)] * (1+ Δx /x )^a[sin(1/x)] ^[sin(1/(x+Δx) / sin(1/x) ]
- x^a[sin(1/x)] } / Δx
∵ Δx -> 0, ∴1+ Δx /x -> 1,
= { x^a[sin(1/x)] ^[sin(1/(x+Δx) / sin(1/x)] * 1 - x^a[sin(1/x)] } / Δx
= x^a[sin(1/x)] * { x^a[sin(1/x)] ^[sin(1/(x+Δx) / sin(1/x)] - 1} / Δx
= x^a[sin(1/x)] * { x^a[sin(1/(x+Δx)] - 1} / Δx
后面楼主自己接着算吧