March | 2017 | zeyoucui

Since d/dx sinx = cosx, sinx=x-1/3!x^3+1/5!x^5-1/7!x^7…(-1)^k/(2k+1)!x^(2k+1) and we know cosx=1-1/2!x^2+1/4!x^4-…+(-1)^k/2k!x^2k. So the derivative of Maclaurin series of sinx equals to cosx
ln(x+1)=x-x^2/2+x^3/3…+(-1)^n/n+1 x^n+1. So we can ger derivative of ln(x+1) is 1-x+x^2-x^3+…+(-1)^nx^n. Since d/dxln(x+1)=1/x+1. We can find the derivative of Maclaurin series for ln(x+1)equals to 1/x+1.
sin(x^2)=x^2-1/3!x^^+2/5!+(-1)^k4k+2/2k!x^4k+1, d/dx sin(x^2)=2xcos(x^2) We can find the derivative of Maclaurin series for sin(x^2)equals to 2xcos(x^2).
for cosx=∑（-1）^n/2n!x^2n, we know d/dx cosx= -sinx. . We can find the derivative of Maclaurin series for cosx equals to -sinx.

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