first
\[\begin{align} \Phi(x)=\int_{0}^{x^{2}} \sin t d t, \enspace \Phi^{\prime}(x)=? \\ \\ 设: u=x^{2} \\ \\ \text { 则: } G(u)=\int_{0}^{u} \sin t d t=\Phi(x) \\ \\ 链式法则: G^{\prime}(u)=G^{\prime}(u) \cdot(u)^{\prime} \\ \\ 微积分基本公式: G^{\prime}(u)=\left(\int_{0}^{u} \sin t d t\right)^{\prime}=\sin (u)=\sin x^{2} \\ \\ (u)^{\prime}=\left(x^{2}\right)^{\prime}=2 x \\ \\ \therefore \Phi^{\prime}(x)=\left(\int_{0}^{x} \sin t d t\right)^{\prime}=2 x \sin x^{2} \end{align} \]second
\[\begin{eqnarray} \lim _{x \rightarrow 0} \frac{\int_{0}^{x} \cos t^{2} d t}{x}=? \\ \\ \because \lim _{x \rightarrow 0} \int_{0}^{x} \cos t^{2} d t=0 \\ \\ \therefore 题目是\frac{0}{0} 型极限，\text { 适用于洛必达法则 } \\ \\ \lim _{x \rightarrow 0} \frac{\int_{0}^{x} \cos t^{2} d t}{x}=\lim _{x \rightarrow 0} \frac{\left(\int_{0}^{x} \cos t^{2} d t\right)^{\prime}}{(x)^{\prime}} \\ \\ \left(\int_{0}^{x} \cos ^{2}t d t\right)^{\prime}=\cos x^{2} \\ \\ \Rightarrow \lim _{x \rightarrow 0} \frac{\cos x^{2}}{1}=\lim _{x \rightarrow 0} \cos x^{2}=1 \end{eqnarray} \]标签：prime,cos,int,积分,lim,例题,sin,rightarrow From： https://www.cnblogs.com/Preparing/p/18288870