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]=1\cdot (e^{x^{2}-x})-0\cdot (e^{3^{2}-3})=e^{x^{2}-x}} Therefore, g ′ ( x ) = e x 2 − x {\displaystyle {\text{Therefore, }}g'(x)=e^{x^{2}-x}}
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![{\displaystyle {\frac {d}{dx}}\left[g(x)\right]={\frac {d}{dx}}\left[\int _{3}^{x}e^{t^{2}-t}dt\right]=1\cdot (e^{x^{2}-x})-0\cdot (e^{3^{2}-3})=e^{x^{2}-x}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/67f74b496dc8f55f072918222a4f5a367ad16a8a)
