find the derivatives of the following functions with respect to x from first principles
y=ln5x
Differentiation
2011-03-06 11:28 pm
回答 (3)
2011-03-06 11:52 pm
✔ 最佳答案
dy/dx= lim(Δx->0) {ln[5(x + Δx)] - ln5x}/Δx
= lim(Δx->0) {ln[5(x + Δx)/5x]}/Δx
= lim(Δx->0) {ln[1 + Δx/5x]}/Δx
= lim(Δx->0) 1/(5x) {ln[1 + Δx/5x]}/(Δx/5x)
= lim(Δx->0) 1/(5x) {ln[1 + Δx/5x]}^[1/(Δx/5x)]
= lim(Δx->0) 1/(5x) {ln[1 + 1/(5x/Δx)]}^(5x/Δx)
= lim(n->infinity) 1/(5x) {ln[1 + 1/n]}^n
= 1/(5x) lne
= 1/(5x)
2011-03-07 11:19:29 補充：
sorry。
= lim(Δx->0) {ln[1 + Δx/5x]}/Δx
這行的5應該是沒有的。以下那些5x換回x就可以了。答案是1/x
2011-03-07 7:35 am
你個答案錯左喎．．．我算計啦ｅ家～
2011-03-07 12:14 am
dy/dx
= lim(Δx->0) {ln[5(x + Δx)] - ln5x}/Δx
= lim(Δx->0) {ln[5(x + Δx)/5x]}/Δx
= lim(Δx->0) {ln[1 + Δx/5x]}/Δx
= lim(Δx->0) 1/(5x) {ln[1 + Δx/5x]}/(Δx/5x)
= lim(Δx->0) 1/(5x) {ln[1 + Δx/5x]}^[1/(Δx/5x)]
= lim(Δx->0) 1/(5x) {ln[1 + 1/(5x/Δx)]}^(5x/Δx)
= lim(n->infinity) 1/(5x) {ln[1 + 1/n]}^n
= 1/(5x) lne
= 1/(5x)
= lim(Δx->0) {ln[5(x + Δx)] - ln5x}/Δx
= lim(Δx->0) {ln[5(x + Δx)/5x]}/Δx
= lim(Δx->0) {ln[1 + Δx/5x]}/Δx
= lim(Δx->0) 1/(5x) {ln[1 + Δx/5x]}/(Δx/5x)
= lim(Δx->0) 1/(5x) {ln[1 + Δx/5x]}^[1/(Δx/5x)]
= lim(Δx->0) 1/(5x) {ln[1 + 1/(5x/Δx)]}^(5x/Δx)
= lim(n->infinity) 1/(5x) {ln[1 + 1/n]}^n
= 1/(5x) lne
= 1/(5x)
參考: up
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