Tìm đạo hàm: ` Sin^{2}x + Sin^{3}x + Sin^{4}x + ... + Sin^{22}x `

$f(x)=\sin^2x+\sin^23x+...+\sin^222x$

Tổng $f(x)$ là tổng 21 số hạng đầu của CSN $u_1=\sin^2x; q=\sin x$.

$\Rightarrow f(x)=\dfrac{\sin^2x(1-\sin^{21}x)}{1-\sin x}$

$=\dfrac{\sin^2x-\sin^2x.\sin^{21}x}{1-\sin x}$

$=\dfrac{\sin^2x-\sin^{23}x}{1-\sin x}$

$f'(x)=\dfrac{(\sin^2x-\sin^{23}x)'(1-\sin x)- (\sin^2x-\sin^{23}x)(1-\sin x)'}{(1-\sin x)^2}$

Xét tử:

$=[(\sin^2x)'-(\sin^{23}x)'](1-\sin x)-(\sin^2x-\sin^{23}x)(-\sin x)'$

$=[2\sin x(\sin x)'- 23\sin^{22}x(\sin x)'](1-\sin x)+(\sin^2x-\sin^{23}x)\cos x$

$=(2\sin x\cos x-23\sin^{22}x\cos x)(1-\sin x)+\cos x(\sin^2x-\sin^{23}x)$

$=\sin2x-\sin2x\sin x-23\sin^{22}x\cos x+23\sin^{23}x\cos x+\sin^2x\cos x-\sin^{23}x\cos x$

Vậy:

$f'(x)=\dfrac{\sin2x-\sin2x\sin x-23\sin^{22}x\cos x+22\sin^{23}x\cos x+\sin^2x\cos x}{(1-\sin x)^2}$

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