first
\[\begin{align}
\Phi(x)=\int_{0}^{x^{2}} \sin t d t, \enspace \Phi^{\prime}(x)=?
\\ \\
設: u=x^{2}
\\ \\
\text { 則: } G(u)=\int_{0}^{u} \sin t d t=\Phi(x)
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鏈式法則: G^{\prime}(u)=G^{\prime}(u) \cdot(u)^{\prime}
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微積分基本公式: G^{\prime}(u)=\left(\int_{0}^{u} \sin t d t\right)^{\prime}=\sin (u)=\sin x^{2}
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(u)^{\prime}=\left(x^{2}\right)^{\prime}=2 x
\\ \\
\therefore
\Phi^{\prime}(x)=\left(\int_{0}^{x} \sin t d t\right)^{\prime}=2 x \sin x^{2}
\end{align}
\]
second
\[\begin{eqnarray}
\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \cos t^{2} d t}{x}=?
\\ \\
\because \lim _{x \rightarrow 0} \int_{0}^{x} \cos t^{2} d t=0
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\therefore 題目是\frac{0}{0} 型極限，\text { 適用於洛必達法則 }
\\ \\
\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \cos t^{2} d t}{x}=\lim _{x \rightarrow 0} \frac{\left(\int_{0}^{x} \cos t^{2} d t\right)^{\prime}}{(x)^{\prime}}
\\ \\
\left(\int_{0}^{x} \cos ^{2}t d t\right)^{\prime}=\cos x^{2}
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\Rightarrow \lim _{x \rightarrow 0} \frac{\cos x^{2}}{1}=\lim _{x \rightarrow 0} \cos x^{2}=1
\end{eqnarray}
\]