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Someone asked me about the following exercise and I tried my best to help them but couldn't understand this. Given:

$$f(x) = \int_2^{-x+3} e^{t^2} dt$$

According to wolframalpha, the derivative is:

$$f'(x) = -e^{(3-x)^2}$$

But I cannot for the life of me figure out where the negation is coming from. I kinda understand that: $$F'(x) = \frac{\partial}{\partial x} \left( \int_a^{x} f(t) dt \right) = f(x)$$

So according to that, if I put $-x+3$ for that x, I'd expect to get $f'(x)=e^{(3-x)^2}$.

What exactly am I missing? I'm sure it's something trivial, but I just can't figure it out and googling didn't help a bit.

Ian Rehwinkel
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    Hint: The chain rule ... – Martin R Feb 15 '22 at 20:13
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    Notice that $f(x) = g(-x + 3)$ with : $$g(x) = \int_2^x e^{t^2} \mathrm{d}t$$ – Essaidi Feb 15 '22 at 20:15
  • Also: https://math.stackexchange.com/q/83317/42969, https://math.stackexchange.com/q/1768396/42969 – Martin R Feb 15 '22 at 20:15
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    LaTeX/MathJax hint: you probably wanted to use \partial instead of \delta. Please compare rendering \delta / \delta x → $\delta / \delta x$ to \partial / \partial x → $\partial / \partial x$. – CiaPan Feb 15 '22 at 20:19
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    The chain rule is the proper way to get the derivative formula including negative sign. But to just see why it will be negative, notice the bounds are in usual order (or identical) if $2 \leq -x+3 \iff x \leq 1$. And the upper limit of integration increases when $x$ decreases: $0 = f(1) < f(0) < f(-1) < \cdots$ – aschepler Feb 15 '22 at 20:24
  • It's the chain rule; $\frac{d}{dx} F[-x+3]=F'[-x+3] \cdot \frac{d}{dx}[-x+3]=-F'[-x+3]$. – Ian Feb 15 '22 at 20:13

1 Answers1

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As I was kindly suggested, the chain rule is behind this. I was thinking the integral rule I mentioned would just work for $-x+3$ aswell.

Therefore the solution is as follows: $$f(x) = \int_{2}^{-x+3} e^{t^2}dt = g(h(x))$$ $$g(x) = \int_{2}^{x} e^{t^2}dt$$ $$g'(x) = e^{t^2}$$ $$h(x) = -x+3$$ $$h'(x) = -1$$ $$f'(x) = (g(h(x)))' = g'(h(x))\cdot h'(x) = e^{h(x)^2} \cdot -1 = -e^{(-x+3)^2}$$

Thanks for the hints and tips, I really should've gotten this myself.

Ian Rehwinkel
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