Đáp án:
$A.\ 1 + x + x^2 +\dfrac{5x^3}{6}+ 0(x^3)$
Giải thích các bước giải:
$\quad y = f(x)=\dfrac{1}{1-\sin x}$
Ta có:
$+)\quad f(0)= 1$
$+)\quad f'(x)= \dfrac{\cos x}{(1-\sin x)^2}\Rightarrow f'(0)= 1$
$+)\quad f''(x)= \dfrac{2\cos^2x -\sin x + \sin^2}{(1-\sin x)^3}\Rightarrow f''(0)= 2$
$+)\quad f'''(x)= \dfrac{\cos x(5\sin^2x-4\sin x + 6\cos^2x -1)}{(1-\sin x)^4}\Rightarrow f'''(0)= 5$
Ta được:
$f(x)= f(0) + \dfrac{f'(0)}{1!}x + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(0)}{3!}x^3 + 0(x^3)$
$\Leftrightarrow \dfrac{1}{1-\sin x}= 1 + x + x^2 +\dfrac{5x^3}{6}+ 0(x^3)$
