Suppose $G:[-1,1]\to\Bbb R$, $G(x):=\int_{-5}^{2x^4+x^2+1}{e^{-t^4}}$, $x \in [-1,1]$. I want to find a formula for the derivative without the integral symbol.

Let $G:[-1,1]\to\Bbb R$, $G(x):= \int_{-5}^{2x^4+x^2+1}{e^{-t^4}} \;dt$, for $x \in [-1,1].$

Solution: Let $C:[-5,4]\to\Bbb R$ be given by $C(x):=\int_{-5}^{x}{e^{-t^4}}$ for all $x \in [1,4]$. The function $c:[-1,1]\to\Bbb R$ given by $c(t)=e^{-t^4}$ for all $t \in [-1,1]$. The function $c$ is continuous as a composition of $t\mapsto e^t$ and $t\mapsto -t^4$ where first is an elementary function and second is a polynomial.

Hence, apply Fundamental Theorem of Calculus for continuous functions, which implies that $C$ is differentiable with $C'(x) = c(x) = e^{-x^4}$ for all $x\in \Bbb R$.

Now, observe that $G(x) = C(2x^4+x^2+1)$, so $G=C\circ g$, where $g(x):=2x^4+x^2+1$ is differentiable as a polynomial. Also, $C$ is differentiable by previous discussion using Fundamental Theorem of Calculus, so the chain rule implies that $G$ is differentiable with derivative:

$G'(x) = C'(g(x))g'(x) ({e^{-(2x^4+x^2+1)^4}})(8x^3+2x)$ for all $x \in [-1,1]$ as required. $\square$

I am really confused about the domain. I know the upper endpoint of the integral is for $[1,4]$. Can someone please help me and edit my solution where the domain / $x$-values are wrong. I think I am getting confused between the domain of the function and values of the integral. It's really annoying me because I can't seem to understand for two weeks. Can I set all $x\in\cdots$ and $[]$ to $\Bbb R$??? I don't think I can because in the question $G$ is defined for $[-1,1]$ where $x \in [-1,1]$.

My definition for the Fundamental Theorem of Calculus. If $f\colon[a,b] \to \Bbb R$ is continuous and $F\colon [a,b] \to \Bbb R$ is given by $F(x):=\int_{a}^{x}f$ for all $x\in[a,b]$, then $F$ is differentiable with derivative $F'(x)=f(x)$ for all $x \in [a,b]$.

  • $\begingroup$ @JonathanZsupportsMonicaC Edited. It should be $G$ the whole way through with the lower limit being $-5$. $\endgroup$
    – user1071088
    Commented Nov 27, 2022 at 20:17
  • $\begingroup$ Why do you care about the domain ? You can extend the function $G$ to the function $\tilde{G}(x)$ defined on the whole real line by the same formula. Then $(\tilde{G})'(x)=e^{-x^4}(8x^3+2x)$ for any $x.$ Hence $G'(x)=(\tilde{G})'(x)$ for $-1<x<1$ and $G'_-(1)=(\tilde{G})'(1),$ $ G'_+(-1)=(\tilde{G})'(-1).$ $\endgroup$ Commented Nov 27, 2022 at 21:36
  • $\begingroup$ @RyszardSzwarc Because look at my definition for the fundamental theorem of calculus which I must use in the question. The question already defines the domain of $G$ to be $[-1,1]$. I must use this $\endgroup$
    – user1071088
    Commented Nov 27, 2022 at 21:50
  • 1
    $\begingroup$ The function $C$ in your answer can be defined for every $x.$ Then $G(x)=C(2x^4+x^2+1)$ for $-1\le x\le 1.$ Let $H(x)=C(2x^4+x^2+1)$ for $x\in\mathbb{R}.$ Then $G(x)=H(x)$ for $-1\le x\le 1.$ Hence $G'(x)=H'(x)$ for $-1<x<1.$ By the fundamental theorem of calculus $H'(x)=e^{-x^2}(8x^3+2x).$ Hence $G'(x)=e^{-x^2}(8x^3+2x)$ for $-1<x<1.$ $\endgroup$ Commented Nov 27, 2022 at 22:02
  • $\begingroup$ @RyszardSzwarc okay can please copy my post with your edits for a complete answer so I can accept it. It would also make more sense to me (also you used strict inequalities sometimes) $\endgroup$
    – user1071088
    Commented Nov 27, 2022 at 23:30

1 Answer 1


The function $G:[-1,1]\to \mathbb{R}$ is defined by the formula $$G(x)=\int\limits_{-5}^{2x^4+x^2+1}e^{-t^4}\,dt$$ The function $G$ is not defined off the interval $[-1,1],$ or it may defined there by the same or by another formula. It will be irrelevant for the solution.

As we are dealing with the closed interval, we may calculate $G'(x)$ for $-1<x<1$ and $G'_+(-1)$ and $G'_-(1).$ We have to use one sided derivatives at the endpoints of the interval, as we have no information on the behaviour of $G$ for $x<-1$ and for $x>1.$

Consider the function $H:\mathbb{R}\to \mathbb{R}$ defined by the same formula as $G$ $$H(x)=\int\limits_{-5}^{2x^4+x^2+1}e^{-t^4}\,dt$$ but with no restriction on $x.$ By the fundamental theorem of calculus and the chain rule, we get $$H'(x)=e^{-(2x^4+x^2+1)^4}(8x^3+2x)$$ Clearly $G(x)=H(x),$ for $|x|\le 1.$ Therefore $$G'(x)=H'(x),\ |x|<1,\ G'_+(-1)=H'(-1),\quad G'_-(1)=H'(1)$$ Thus $$G'(x)=e^{-(2x^4+x^2+1)^4}(8x^3+2x),\ -1<x<1$$ and $G'_+(-1)=-10e^{-64}$,$\ $ $G'_-(1)=10e^{-64}$


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