运用含参变量函数积分求导

青青子衿

\begin{gather*}
\dfrac{\mathrm{d}}{\mathrm{d}v}\left[\int_{b\left(v\right)}^{a\left(v\right)}w\left(u,v\right)\mathrm{d}u\right]\\
=\int_{b\left(v\right)}^{a\left(v\right)}\dfrac{\partial\,w\left(u,v\right)}{\partial\,v}\mathrm{d}u
+w\big[a\left(v\right),v\big]\dfrac{\mathrm{d}a\left(v\right)}{\mathrm{d}v}
-w\big[b\left(v\right),v\big]\dfrac{\mathrm{d}b\left(v\right)}{\mathrm{d}v}
\end{gather*}

青青子衿

\begin{align*}
u&\quad\overline{\hspace{1cm}}\quad t\\
v&\quad\overline{\hspace{1cm}}\quad x\\
w\left(u,v\right)&\quad\overline{\hspace{1cm}}\quad t\cdot\,f\left(x^2-t^2\right)\\
a\left(v\right)&\quad\overline{\hspace{1cm}}\quad x\\
b\left(v\right)&\quad\overline{\hspace{1cm}}\quad 0\\
\end{align*}

\begin{align*}
&\dfrac{\mathrm{d}}{\mathrm{d}x}\left[\int_{0}^{x}t\cdot\,\!f\left(x^2-t^2\right)\mathrm{d}t\right]\\
=&\int_{0}^{x}\dfrac{\partial\left[t\cdot\,\!f\left(x^2-t^2\right)\right]}{\partial\,x}\mathrm{d}t
+x\cdot\,\!f\big(x^2-x^2\big)\dfrac{\mathrm{d}x}{\mathrm{d}x}
-0\cdot\,\!f\big(x^2\big)\dfrac{\mathrm{d}0}{\mathrm{d}x}\\
=&\int_{0}^{x}\dfrac{\partial\left[t\cdot\,\!f\left(x^2-t^2\right)\right]}{\partial\,x}\mathrm{d}t
+x\cdot\,\!f\big(0\big) \\
=&\int_{0}^{x}t\cdot\,\!f_{\overset{\,}{x}}\left(x^2-t^2\right)\cdot\,\!2x\,\mathrm{d}t
+x\cdot\,\!f\big(0\big)\\
=&\int_{0}^{x}\left(-x\right)\cdot\,\!f_{\overset{\,}{t}}\left(x^2-t^2\right)\cdot\,\!\left(-2t\right)\,\mathrm{d}t
+x\cdot\,\!f\big(0\big)\\
=&\left(-x\right)\int_{0}^{x}f_{\overset{\,}{t}}\left(x^2-t^2\right)\cdot\,\!\left(-2t\right)\,\mathrm{d}t
+x\cdot\,\!f\big(0\big)\\
=&\left(-x\right)\left.\Big[f\left(x^2-t^2\right)\Big]\vphantom{\overset{1}{1}}\right|_{0}^{x}
+x\cdot\,\!f\big(0\big)\\
=&\left(-x\right)f\left(0\right)+x\cdot\,\!f\left(x^2\right)
+x\cdot\,\!f\big(0\big)\\
=&x\cdot\,\!f\left(x^2\right)
\end{align*}