1.试求函数 =(int )_(0)^xsin tdt 当 x=0 及 =dfrac (pi )(4) 时的导数.

求下列极限: lim _(xarrow 0)[ dfrac ({int )_(0)^xsqrt (1+{t)^2}dt}(x)+dfrac ({int )_(0

设函数 f(x)= ^2)(int )_(0)^xtan xdt xlt 0 1 x=0 dfrac (1)({x)^2}(int )_(0)^xsin

3.求由 (int )_(0)^y(e)^tdt+(int )_(0)^xcos tdt=0 所确定的隐函数对x的导数 dfrac (dy)(dx).

3.求由 (int )_(0)^y(e)^tdt+(int )_(0)^xcos tdt=0 所决定的隐函数对x的导数 dfrac (dy)(dx).

3.求由 (int )_(0)^y(e)^tdt+(int )_(0)^xcos tdt=0 所确定的隐函数对x的导数 dfrac (dy)(dx) .

9.求下列函数在给定点处的导数值:-|||-(1) =cos xsin x ,求 (|)_(x)=dfrac (pi )(6) 和 (|)_(x)=dfrac

定积分(int )_(0)^dfrac (pi {2)}(e)^xsin xdx的值是(int )_(0)^dfrac (pi {2)}(e)^xsin xdx

(int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx=( ) .(int )_(0)^dfrac (pi {4)}dfr

(int )_(0)^dfrac (pi {2)}xsin xdx= ()-|||-

极限lim _(xarrow {0)^+}dfrac ({int )_(0)^(x^4)dt(int )_(0)^xsin dfrac (t)(u)du}(1-