5.3 The Fundamental Theorem of Calculus/8

$g(x)=\int _{3}^{x}e^{t^{2}-t}dt{\frac {d}{dx}}\left[g(x)\right]={\frac {d}{dx}}\left[\int _{3}^{x}e^{t^{2}-t}dt\right]=1e^{x^{2}-x}-0e^{3^{2}-3}=e^{x^{2}-x}{\text{Therefore, }}g'(x)=e^{x^{2}-x}$

${\begin{aligned}u&={\tfrac {1}{\sqrt {2}}}(x+y)\qquad &x&={\tfrac {1}{\sqrt {2}}}(u+v)\\[0.6ex]v&={\tfrac {1}{\sqrt {2}}}(x-y)\qquad &y&={\tfrac {1}{\sqrt {2}}}(u-v)\end{aligned}}$