g ( x ) = ∫ 3 x e t 2 − t d t {\displaystyle g(x)=\int _{3}^{x}e^{t^{2}-t}dt} d d x [ g ( x ) ] = d d x [ ∫ 3 x e t 2 − t d t ] = 1 e x 2 − x − 0 e 3 2 − 3 = e x 2 − x {\displaystyle {\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}} Therefore, g ′ ( x ) = e x 2 − x {\displaystyle {\text{Therefore, }}g'(x)=e^{x^{2}-x}}
\end{align} </math>
u = 1 2 ( x + y ) x = 1 2 ( u + v ) v = 1 2 ( x − y ) y = 1 2 ( u − v ) {\displaystyle {\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}}}
![{\displaystyle {\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}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/01578fa5504adae02e29e674ceeaa68e9f4b7884)
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/4e0308a3a70f9d0b044c9dd0541c27a130ddbeac)