# Math Help - [SOLVED] Derivatives of integrals

1. ## [SOLVED] Derivatives of integrals

I "need" you, MHF.
I would like you to verify the answers I got (I can't trust Mathematica) and help me with the 2 last problems.
Find the derivatives of the following integrals :
1) $F(x)=\int_x^b \frac{dt}{1+t^2+sin^2(t)}$. I got $F'(x)=-\frac{1}{1+x^2+sin^2x}$.
2) $F(x)=\int_3^{\int_1^x sin^3(t)dt} \frac{du}{1+sin^6(u)+u^2}$. I got $F'(x)=\frac{sin^3x}{1+sin^6(\int_1^x sin^3(x)dt)+(\int_1^x sin^3(x)dt)^2}$.
3) $F(x)=\int_a^x xf(t)dt$. I don't know how to start this one.
4) $F(x)=sin(\int_0^x sin(\int_0^y sin^3(t)dt)dy)$. I am not able to do this one. But I have started something : I wrote $F$ as a composition of 3 functions, that is $f \circ g \circ h$.
$f(g)=sin(g)$.
$g(h)=\int_0^x sin(h)dy$.
$h(y)=\int_0^y sin^3(t)dt$.
$f'(g)=cos(g)$
$g'(h)=$... I couldn't find it. I guess I'm already wrong writing $g(h)$. All those variables confuse me a lot!
$h'(y)=sin^3(y)$.
I wanted to use the chain rule for this one. $F'(x)$ would be $(f' \circ g *g' \circ h * h')(x)$. I'm not even sure if this is right.
Also, I may post one more derivative of an integral.

2. Hi arbolis

Originally Posted by arbolis
Find the derivatives of the following integrals :
1) $F(x)=\int_x^b \frac{dt}{1+t^2+sin^2(t)}$. I got $F'(x)=-\frac{1}{1+x^2+sin^2x}$.
OK
2) $F(x)=\int_3^{\int_1^x sin^3(t)dt} \frac{du}{1+sin^6(u)+u^2}$. I got $F'(x)=\frac{sin^3x}{1+sin^6(\int_1^x sin^3({\color{red}t})dt)+(\int_1^x sin^3({\color{red}t})dt)^2}$.
OK

(By the way, $\int_1^x\sin^3t\,\mathrm{d}t$ can be computed using $\sin^3x=\sin ^2x \sin x=\sin x -\cos ^2x \sin x$)

3) $F(x)=\int_a^x xf(t)dt$. I don't know how to start this one.
The integration is done with respect to $t$ so one can write $F(x)=x\int_a^xf(t)\,\mathrm{d}t$ and differentiate using the product rule. ( $u(x)=x,\,v(x)=\int_a^xf(t)\,\mathrm{d}t$)

4) $F(x)=sin(\int_0^x sin(\int_0^y sin^3(t)dt)dy)$. I am not able to do this one. But I have started something : I wrote $F$ as a composition of 3 functions, that is $f \circ g \circ h$.
$f(g)=sin(g)$.
$g(h)=\int_0^x sin(h)dy$.
$h(y)=\int_0^y sin^3(t)dt$.
$f'(g)=cos(g)$
$g'(h)=$... I couldn't find it. I guess I'm already wrong writing $g(h)$. All those variables confuse me a lot!
$h'(y)=sin^3(y)$.
I wanted to use the chain rule for this one. $F'(x)$ would be $(f' \circ g *g' \circ h * h')(x)$. I'm not even sure if this is right.
$F(x)=\sin\left(\int_0^x \sin\left[\int_0^y \sin^3 t\, \mathrm{d}t\right]\,\mathrm{d}y\right)$

Using composite functions is the right idea but you can't define $g(h)=\int_0^x \sin h \,\mathrm{d}y$ since the integration is done with respect to $y$ and $h$ depends on $y$. You'll have to use $g(x)= \int_0^x \sin\left[\int_0^y \sin^3(t)\,\mathrm{d}t\right]\,\mathrm{d}y$. This gives $F(x)=f\circ g(x)$ hence $F'(x)=g'(x)\cdot f'\circ g(x)$. (chain rule)

You've already found $f'$ but what is $g'$ ? Remember that $\frac{\mathrm{d}}{\mathrm{d}x}\left[ \int_0^x u(y)\,\mathrm{d}y\right]=u(x)$. What does $u(y)$ correspond to here ?

3. Hello Écureuil volant!
Oops, sorry for the 2), I mistyped, so I got the right answer (on my sheet).
For the 3),
The integration is done with respect to so one can write and differentiate using the product rule. ()
Thanks! I was very confused. I understand better now. So $F'(x)=\int_a^xf(t)\,\mathrm{d}t+x^2f(t)$.
For the 4),
What does correspond to here ?
It corresponds to $sin \big (\int_0^y \sin^3 t\, \mathrm{d}t\big )$.
So $F'(x)=cos \left( \int_0^x \sin\left[\int_0^y \sin^3 (t) dt\right] dy\right) \cdot \sin \big( \int_0^x \sin^3(t) dt \big)$? Or I am confused again about the variables?

4. And finally the last one : $F(x)=\int_a^b \frac{x}{1+t^2+sin^2t} dt$. I considered x as a constant when under the integral since it's a dt one, as you just teached me. I got $F'(x)=\int_a^b \frac{dt}{1+t^2+sin^2t}$. Hope it's right.

5. Originally Posted by arbolis
Hello Écureuil volant!

For the 3), Thanks! I was very confused. I understand better now. So $F'(x)=\int_a^xf(t)\,\mathrm{d}t+x^2f(t)$.
You meant $F'(x)=\int_a^x f(t)\,\mathrm{d}t+{\color{red}xf(x)}$ didn't you ?
For the 4), It corresponds to $sin \big (\int_0^y \sin^3 t\, \mathrm{d}t\big )$.
So $F'(x)=cos \left( \int_0^x \sin\left[\int_0^y \sin^3 (t) dt\right] dy\right) \cdot \sin \big( \int_0^x \sin^3(t) dt \big)$?
That's it

EDIT :
Originally Posted by arbolis
And finally the last one [...] Hope it's right.
It is.

6. You meant didn't you ?
Yes sorry about that, I did it mentally and probably too fast as we can see. Thanks a lot for your precious help/time.