# Let $f$ be a cont. on $\mathbb{R}$ and define $G(x)=\int_0^{\sin (x)}f(t) dt$. Show that $G$ is differentiable on $\Bbb{R}$ and compute $G'$.

Let $f$ be a continuous function on $\mathbb{R}$ and define $$G(x)=\int_0^{\sin (x)}f(t) dt$$ Show that $G$ is differentiable on $\mathbb{R}$ and compute $G'$.

This is an exercise from Elementary Analysis: The Theory of Calculus by Kenneth A. Ross.

My first idea is the following. Let $x\in \Bbb R$. Define $F(x):=\int_0^{x}f(t) dt$. And prove that $G:=F\circ \sin$ is differentiable at $x$ using:

It is clear that $\sin(x)$ is differentiable at $x$. So I need to proof that $F$ is differentiable at $y:=\sin(x)$. I was trying to prove that using:

I'm not sure if I can use this theorem. Obviously, $f$ sattisfies the conditions. But as $y=\sin(x) \in [-1,1]$, I dont see how I could choose $a,b$ in theorem 34.3, such that it fits this case. It seems clear that I should choose $a:=0$, and if $y>0$ I could choose $b:=1$, but if $y<0$, I can't find any $b$ such that $y\in[a,b]$.

I wanted to know if I'm heading in the right direction, so I looked at the official solution manual. What they are doing there seems complete nonsense to me:

I don't see why they are doing what they are doing. They end the proof with $G$ is continuous. Why is that result needed ? And isn't $f$ confused for (some undefined) $F$ in the beginning?

As this proof didn't satisfy me, I looked further, and I found this proof:

This proof seems completely solid to me. But it is much less intuitive (for me). I wouldn't have come with this proof myself.

My question are:

1. Was I heading in the right direction with my first idea ? Or is it not possible to apply theorem 34.3 in this kind of way ?
2. Am I right that the official solution manual is complete nonsense, or am I missing something ?
3. Is this last proof I found correct, and do you think the author of the book has this kind of proof in mind when he wrote this exercise ? Or do you think the author was expecting some other kind of proof ?

Edit: For the bounty I would like to see a rigorous prove of why $G$ is differentiable on all $x\in \Bbb R$ using only theorems that are proven in Elementary Analysis: The Theory of Calculus by Kenneth A. Ross.

• Oh, haha, will I didn't meant to add a 300 point bounty, don't know what happened, I thought I had selected a 50 point bounty. Anyway, you better give some well thought answer for 300 points ;) – Kasper Oct 8 '14 at 0:27

We have \begin{align} G(x) &=F(\sin(x))\\ &=\int_0^{\sin(x)}f(t)\,\mathrm{d}t\tag{1} \end{align} where \begin{align} F(u) &=\int_0^uf(t)\,\mathrm{d}t\\ &=\underbrace{\int_{-1}^uf(t)\,\mathrm{d}t}_{\substack{\text{differentiable}\\\text{for u\ge-1}}}-\underbrace{\int_{-1}^0f(t)\,\mathrm{d}t}_\text{constant}\tag{2} \end{align} The Fundamental Theorem of Calculus says $$F'(u)=f(u)\tag{3}$$ The Chain Rule says that \begin{align} G'(x) &=F'(\sin(x))\cos(x)\\ &=f(\sin(x))\cos(x)\tag{4} \end{align}

Technical Point: To apply Theorem 28.4 to Theorem 34.3 without any further work, we should choose the lower bound of integration in $(2)$ to be lower than $-1$.

However, with a bit of extra work, we can show that $f$ need only be continuous on $[-1,1]$. The only problem arises when computing $F'(\pm1)$. Since $F(u)$ will only see $u\in[-1,1]$, we only need to consider the one-sided derivative at $\pm1$.

Suppose $u_0=\pm1$, then for $u\in(-1,1)$ we can apply the Mean Value Theorem to find a $\xi$ between $u_0$ and $u$, hence in $(-1,1)$, so that \begin{align} \frac{F(u)-F(u_0)}{u-u_0} &=F'(\xi)\\ &=f(\xi)\tag{5} \end{align} Therefore, since $f$ is continuous on $[-1,1]$, we have the one-sided derivative \begin{align} F'(u_0) &=\lim_{u\to u_0}\frac{F(u)-F(u_0)}{u-u_0}\\ &=\lim_{\xi\to u_0}f(\xi)\\ &=f(u_0)\tag{6} \end{align}

1. You are thinking correctly about applying the Fundamental Theorem of Calculus and the Chain Rule.

2. I am not sure why they are showing that $G$ is continuous. Since they have shown its derivative exists, it is already continuous.

3. The handwritten proof looks okay. It is a different approach, but it is valid. You don't need the Chain Rule with that approach.

• Okay, but we need to prove this for all $x\in \Bbb R$. So $u=\sin x \in [-1,1]$. So if you look at theorem 34.3 very litterally, then it seems like you can't use it. You have to choose $a=0$. But $u$ could be less than zero. I thought that may be the reason why the handwritten prove uses this other technique. – Kasper Oct 8 '14 at 1:03
• What "this" do we need to prove that we haven't? The fact that $u\in[-1,1]$ simply means we only need to know $f(t)$ on $[-1,1]$. Please clarify your concern. – robjohn Oct 8 '14 at 1:13
• I mean proving that $G$ is differentiable on all $x\in \Bbb R$. To prove that you need that show that $F$ is differentiable on $u\in[-1,1]$. You say that you can use theorem 34.3 (see my post for an attached picture) for that. But this theorem proves only that $u\in(a,b)$ is differentiable. Where you have chosen $a$ to be $0$ it seems. – Kasper Oct 8 '14 at 1:24
• @Kasper: I now understand your concern with Theorem 34.3, and I have added a bit to show that it still applies. – robjohn Oct 8 '14 at 5:49
• @Kasper: to apply Theorem 34.3 as given, yes. However, I have added an argument to show that we get the same result even if we only know that $f$ is continuous on $[-1,1]$. – robjohn Oct 8 '14 at 11:55