Quay lại đáp án

	3	sửa đáp án
a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$ b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$ c. $y'=3sin^2(2x+1)[sin(2x+1)]'=3sin^2(2x+1)[cos(2x+1).(2x+1)']=6sin^2(2x+1).cos(2x+1)=3sin(4x+2).sin(2x+1)$ a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$ b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$ c. $y'=3sin^2(2x+1)[sin(2x+1)]'=2sin^2(2x+1)[cos(2x+1).(2x+1)']=4sin^2(2x+1).cos(2x+1)=2sin(4x+2).sin(2x+1)$ a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$ b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$ c. $y'=3sin^2(2x+1)[sin(2x+1)]'=3sin^2(2x+1)[cos(2x+1).(2x+1)']=6sin^2(2x+1).cos(2x+1)=3sin(4x+2).sin(2x+1)$

	2	thêm 17 ký tự trong đáp án
a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$ b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$ c. $y'=3sin^2(2x+1)[sin(2x+1)]'=2sin^2(2x+1)[cos(2x+1).(2x+1)']=4sin^2(2x+1).cos(2x+1)=2sin(4x+2).sin(2x+1)$ a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$ b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$ c. $y'=2sin(2x+1)[sin(2x+1)]=2sin(2x+1)[cos(2x+1).(2x+1)']=4sin(2x+1).cos(2x+1)=2sin(4x+2)$ a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$ b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$ c. $y'=3sin^2(2x+1)[sin(2x+1)]'=2sin^2(2x+1)[cos(2x+1).(2x+1)']=4sin^2(2x+1).cos(2x+1)=2sin(4x+2).sin(2x+1)$

	1	tạo mới
a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$ b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$ c. $y'=2sin(2x+1)[sin(2x+1)]=2sin(2x+1)[cos(2x+1).(2x+1)']=4sin(2x+1).cos(2x+1)=2sin(4x+2)$

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