Đáp án:
$\dfrac{1}{1+\sin^2x}= 1 - x^2 + \dfrac{4x^4}{3} + o(x^6)$
Giải thích các bước giải:
$\quad f(x)= \dfrac{1}{1+\sin^2x}$
$+)\quad f(0)= 1$
$+)\quad f'(x)= - \dfrac{sin2x}{(1+\sin^2x)^2}\Rightarrow f'(0)= 0$
$+)\quad f''(x)= \dfrac{4(6\cos2x +\cos4x -3 )}{(cos2x -3)^3}\Rightarrow f''(0)= -2$
$+)\quad f'''(x)= \dfrac{8\sin2x(24\cos2x +\cos4x +7)}{(\cos2x -3)^4} \Rightarrow f'''(0)= 0$
$+)\quad f^{(4)}(x)= -\dfrac{4(-66\cos2x + 320\cos4x + 66\cos6x + \cos8x - 65)}{(\cos2x -3)^5} \Rightarrow f^{(4)}(0)= 32$
$+)\quad f^{(5)}(x)= -\dfrac{8\sin2x(3204\cos2x +2140\cos4x +156\cos6x + cos8x -573)}{(\cos2x -3)^6}\Rightarrow f^{(5)}(0)= 0$
Ta được:
$f(x)= 1 - x^2 + \dfrac{4x^4}{3} + o(x^6)$
