高数高手请进!!

解：1.dy=(arccos(1/x)+lnx)'dx
=[-(-1/x²)/(√(1-(1/x²)))+1/x]dx
=(1/x)[1/√(x²-1)+1]dx;
2.原式=f(x²)*(x²)'-f(lnx)(lnx)'
=2xf(x²)-f(lnx)/x;
3.原式=∫(-1,1)(1-2x²+x^4)dx
=(x-2/3x³+1/5x^5)│(-1,1)
=2(1-2/3+1/5)
=16/15;
4.原式=-1/2∫(0,2)√(4-x²)d(4-x²)
=[(-1/2)(3/2)√(4-x²)³]│(0,2)
=(-1/2)(3/2)(0-8)
=6;
5.∵√(1+sinx)=√(sin²(x/2)+cos²(x/2)+2sin(x/2)cos(x/2))
=√(sin(x/2)+cos(x/2))²
=sin(x/2)+cos(x/2)
∴原式=lim(x->0)[(sin(x/2)+cos(x/2)-cosx)/(x(e^x-1))]
=lim(x->0)[(cos(x/2)/2-sin(x/2)/2+sinx)/(xe^x+e^x-1)]
(0/0型，应用罗比达法则)
=1/0=不存在；（此题可能打错了！）
6.∵x=t+sint,y=e^tcost
∴当x=0，y=1时，t=0.
∴曲线在点(0,1)处切线的斜率k1=y'
=(e^tcost)'/(t+sint)' (t=0时)
=(e^tcost-e^tsint)/(1+cost) (t=0时)
=1/2
曲线在点(0,1)处切线的斜率k2=-1/k1=-2
故曲线在点(0,1)处切线方程是: y=x/2+1
曲线在点(0,1)处法线方程是：y=-2x+1.

如图~~不太清楚见谅~~

打公式费劲啊~

1
dy=y' dx
=(1+√(x²-1))/(x*(√（x²-1）)) dx (括号自己看着办)
第二题还没学到
3
原式=（1-2/3+1/5）-（-1+2/3-1/5）=26/15
...后面不会...