17e0b849d3a4a3b.jpg,计算[img=19x34]17e0ab14a855463.jpg[/img]的实验命令为（ ）. A: syms x; f=diff((1+sin(x)^2)/cos(x),1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2 B: f=diff((1+sinx^2)/cosx,1)f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;f=diff((1+sinx^2)/cosx,1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2

[x^2*sin(1/x^2)]/x的X趋于0的极限,为什么不能用sin(1/x^2)~1/x^2带入.

设\(z = \int_ { { x^2}}^y { { e^t}\sin t} dt\)，则\({z_{xx}=}\) A: \(2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) B: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} - 2{x^2}\cos {x^2}} \right]\) C: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) D: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\cos {x^2} + 2{x^2}\sin {x^2}} \right]\)

$\int {{1 \over {3 + 5\cos x}}} dx = \left( {} \right)$ A: ${1 \over 4}\ln \left| {{{2\cos x + \sin x} \over {2\cos x - \sin x}}} \right| + C$ B: ${1 \over 4}\ln \left| {{{2\cos {x \over 2} + \sin {x \over 2}} \over {2\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ C: $\ln \left| {{{\cos {x \over 2} + \sin {x \over 2}} \over {\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ D: $\ln \left| {{{\cos x + \sin x} \over {\cos x - \sin x}}} \right| + C$

求下列不定积分.[tex=7.286x2.643]28VI4S//fW038PiMAbBHktfj3FfJYocy4+TgcP5gH+6DCjcL5MVe5w4GLCJx2oaC[/tex].腺 由于 $\sin ^{4} x+\cos ^{4} x=\left(\cos ^{2} x-\sin ^{2} x\right)^{2}+2 \sin ^{2} x \cos ^{2} x$$=\cos ^{2} 2 x+\frac{1}{2} \sin ^{2} 2 x$原式 $=\int \frac{\mathrm{d} x}{\cos ^{2} 2 x+\frac{1}{2} \sin ^{2} 2 x}$

函数\(y = \sin {1 \over x}\)的导数为（ ）. A: \({1 \over { { x^2}}}\sin {1 \over x}\) B: \( - {1 \over { { x^2}}}\sin {1 \over x}\) C: \( - {1 \over { { x^2}}}\cos {1 \over x}\) D: \({1 \over { { x^2}}}\cos {1 \over x}\)

$\int {{{x\cos x} \over {{{\sin }^3}x}}} dx = \left( {} \right)$ A: $ - {x \over {2{{\sin }^2}x}} - {1 \over 2}\tan x + C$ B: $ - {x \over {2{{\sin }^2}x}} - {1 \over 2}\cot x + C$ C: $ - {x \over {2{{\cos }^2}x}} - {1 \over 2}\cot x + C$ D: $ - {x \over {2{{\cos }^2}x}} - {1 \over 2}\tan x + C$

求函数[img=107x38]17da6537b12a2e0.png[/img]的导数； ( ) A: 2*x*sin(1/x) - sin(1/x) B: 2xsin(1/x) - cos(1/x) C: 2*x*sin(1/x) - cos(1/x) D: 2*x*cos(1/x) - cos(1/x)

$\int {{{\sin 2x} \over {1 + {{\sin }^4}x}}} {\rm{d}}x = $ A: $\arctan (\sin x) + C$ B: $\arctan ({\sin ^2}x) + C$ C: ${\arctan ^2}(\sin x) + C$ D: $ - {\arctan ^2}(\sin x) + C$

\( \lim \limits_{x \to 0} { { \sqrt {1 + x\sin x} - \cos x} \over { { {\sin }^2}{x \over 2}}} = \)______ 。