![](https://blog.hocmai.vn/wp-content/uploads/2017/07/hot.gif)
![](https://blog.hocmai.vn/wp-content/uploads/2017/07/hot.gif)
Tính đạo hàm bằng định nghĩa:
1) y=sin(2x+4)
2) y=cos(4x+2018)
3) y=tan(5x−2)
1) y=sin(2x+4)
2) y=cos(4x+2018)
3) y=tan(5x−2)
1)[imath]y=sin(2x+4)[/imath]
[imath]\rightarrow y' = sin'(2x+4) = (2x+4)'.cos(2x+4) = [(2x)'+(4)'].cos(2x+4) =2cos(2x+4)[/imath]
2) [imath]y=cos(4x+2018)[/imath]
[imath]\rightarrow y' = cos'(4x+2018) = -(4x+2018)'.sin(4x+2018) = -[(4x)'+(2018)'].sin(4x+2018) = -4.sin(4x+2018)[/imath]
3)[imath]y=tan(5x−2)[/imath]
[imath]\rightarrow y' = tan'(5x-2) = (5x-2)'.\dfrac{1}{cos^2(5x-2)} = \dfrac{5}{cos^2(5x-2)}[/imath]
@chi254 Em làm như này đúng hong ạ :3
[imath]\lim \limits_{x\to 0} \dfrac{\cos x}{x}[/imath] không tồn tại do [imath]\lim \limits_{x\to 0^+} \dfrac{\cos x}{x}=+\infty; \lim \limits_{x\to 0^-} \dfrac{\cos x}{x}=-\infty[/imath]
còn [imath]\lim\limits_{x \to \infty} \dfrac{sinx}{x} = 1[/imath] đúng ko chị ơi :>