# Thread: Proof that G is differentiable on \R & compute G'

1. ## Proof that G is differentiable on \R & compute G'

Hullo,

One more question for the night. I need to prove the following:

Let $\displaystyle f$ be a continuous function on $\displaystyle \mathbb{R}$
and define $\displaystyle G(x)=\int_0^{sinx}\; f(t)dt \backepsilon x \in \mathbb{R}.$

Show that $\displaystyle G$ is differentiable on $\displaystyle \mathbb{R}$ and compute $\displaystyle F'.$

Thanks in advance for the help.

-the Doctor

2. Originally Posted by thedoctor818
Hullo,

One more question for the night. I need to prove the following:

Let $\displaystyle f$ be a continuous function on $\displaystyle \mathbb{R}$
and define $\displaystyle G(x)=\int_0^{sinx}\; f(t)dt \backepsilon x \in \mathbb{R}.$

Show that $\displaystyle G$ is differentiable on $\displaystyle \mathbb{R}$ and compute $\displaystyle F'.$

Thanks in advance for the help.

-the Doctor
I assume you mean G'

Again, the idea is to make the problem more familiar:

Let us define the function $\displaystyle H(x) = \int_0^{x}\; f(t)dt$

This function I assume you know how to differentiate.

Now let's define the function I(x) = sinx, which we also know hoe to differentiate. Now, notice that

$\displaystyle G(x) = H(I(x))$

so $\displaystyle G'(x) = H'(I(x))*I'(x)$...

3. I appreciate your help, but I still am a bit confused. So, should the derivative be: $\displaystyle -sin(x^2)cos(x)?$

I must be having a brain fart, but I am having trouble tonight with the derivative of the basic integral you proposed: $\displaystyle \int_0^{x^2}\;f(t)dt$

I do know that $\displaystyle H'(x_0)=h(x_0)$ for this example, but am not sure how this helps me. Sorry, I have been up to long today & am missing the obvious.

-Michael

4. Originally Posted by thedoctor818
I appreciate your help, but I still am a bit confused. So, should the derivative be: $\displaystyle -sin(x^2)cos(x)?$

I must be having a brain fart, but I am having trouble tonight with the derivative of the basic integral you proposed: $\displaystyle \int_0^{x^2}\;f(t)dt$

I do know that $\displaystyle H'(x_0)=h(x_0)$ for this example, but am not sure how this helps me. Sorry, I have been up to long today & am missing the obvious.

-Michael
$\displaystyle H'(x_0)=f(x_0)$, so $\displaystyle H'(I(x))=f(I(x))=f(sinx)$
$\displaystyle I'(x) = cosx$

also, I'm not sure where you see an $\displaystyle x^2$, I never squared anything.

5. Thanks, I believe that has me straightened out.

-Michael

6. Originally Posted by thedoctor818
Hullo,

One more question for the night. I need to prove the following:

Let $\displaystyle f$ be a continuous function on $\displaystyle \mathbb{R}$
and define $\displaystyle G(x)=\int_0^{sinx}\; f(t)dt \backepsilon x \in \mathbb{R}.$

Show that $\displaystyle G$ is differentiable on $\displaystyle \mathbb{R}$ and compute $\displaystyle F'.$

Thanks in advance for the help.

-the Doctor
As $\displaystyle f$ is continuous it is Riemann integrable on any (finite) interval, and thus put $\displaystyle \int f(x)\,dx=F(x)$ , i.e. $\displaystyle F(x)$ is a primitive function of $\displaystyle f$, so:

$\displaystyle G(x)=\int\limits^{\sin x}_0 f(t)\,dt=F(\sin x)-F(0)\Longrightarrow G'(x)=\cos xF'(\sin x)=\cos xf(\sin x)$ I'm assuming you meant G" and not F'...)

Tonio