# Derivative of an integral question

## Homework Statement

http://imgur.com/hCzGE

## The Attempt at a Solution

I was wondering, why I must take the chain rule of the top, but not the bottom?

should i not take chain rule for both (1-9x)^3 and (1-9x)^2, meaning the two derivatives of 9x (9) cancel out?

Thanks

Last edited:

What does the fundamental theorem state?

if f(t) is continuous for a<t<b and if g(x)= integral a to x for f(x) then g'(x)=f(x) for all a<x<b

lurflurf
Homework Helper
The question is find f'(x) for
$$\mathrm{f}(x)=\int^x_2 \! \left(\frac{1}{2}t^2-1 \right)^{15} \, \mathrm{dt}$$

by the fundamental theorem of calculus

$$\dfrac{d}{dx}\int^x_a \! \mathrm{g}(t) \, \mathrm{dt}=g(x)$$

The differentiation cancels the integration, you need not differentiate the integrant. The chain rule is not needed, but allows for a generalization called Leibniz (integral) rule.

$$\dfrac{d}{dt}\int^{\mathrm{b}(t)}_{\mathrm{a}(t)} \mathrm{f}(x,t) \, \mathrm{dx}=\int^{\mathrm{b}(t)}_{\mathrm{a}(t)} \dfrac{\partial \mathrm{f}}{\partial t} \, \mathrm{dx}+\mathrm{f}(\mathrm{b}(t),t) \mathrm{b} ^ \prime (t)-\mathrm{f}(\mathrm{a}(t),t) \mathrm{a} ^ \prime (t)$$

if f(t) is continuous for a<t<b and if g(x)= integral a to x for f(x) then g'(x)=f(x) for all a<x<b

What are g and f in the problem?

HallsofIvy
Homework Helper

## Homework Statement

http://imgur.com/hCzGE

## The Attempt at a Solution

I was wondering, why I must take the chain rule of the top, but not the bottom?

should i not take chain rule for both (1-9x)^3 and (1-9x)^2, meaning the two derivatives of 9x (9) cancel out?

Thanks
I have no idea what you are talking about! There is NO "chain rule" involved at all.

The problem is to find the derivative of
$$\int_{1- 9x}^1\frac{u^3}{1+u^2}du$$

The only "trick" is that we need to write this as
$$-\int_1^{1- 9x}\frac{u^3}{1+u^2}du$$

So that we can now apply the "fundamental theorem": replace each "u" in the integrand with 1- 9x. The derivative is
$$-\frac{(1- 9x)^3}{1+ (1- 9x)^2}$$

lurflurf
Homework Helper
Is this changing? First it was
$$\mathrm{f}(x)=\int^x_2 \! \left(\frac{1}{2}t^2-1 \right)^{15} \, \mathrm{dt} \\ \text{then it became}\\ \mathrm{y}=\int^1_{1-9x} \! \frac{u^3}{1+u^2} \, \mathrm{du}\\ \text{by the chain rule} \\ \dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-u^\prime \frac{u^3}{1+u^2}$$

SammyS
Staff Emeritus
Homework Helper
Gold Member
Is this changing? First it was
$$\mathrm{f}(x)=\int^x_2 \! \left(\frac{1}{2}t^2-1 \right)^{15} \, \mathrm{dt} \\ \text{then it became}\\ \mathrm{y}=\int^1_{1-9x} \! \frac{u^3}{1+u^2} \, \mathrm{du}\\ \text{by the chain rule} \\ \dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-u^\prime \frac{u^3}{1+u^2}$$
lurflurf,

I wondered about your first reply, post #4 . The problem looked the way it is now posted, back when I first looked. That was before your post in #4. -- You may have seen something different and were composing your response when I first looked at this.

SammyS
Staff Emeritus
Homework Helper
Gold Member
I have no idea what you are talking about! There is NO "chain rule" involved at all.

The problem is to find the derivative of
$$\int_{1- 9x}^1\frac{u^3}{1+u^2}du$$

The only "trick" is that we need to write this as
$$-\int_1^{1- 9x}\frac{u^3}{1+u^2}du$$

So that we can now apply the "fundamental theorem": replace each "u" in the integrand with 1- 9x. The derivative is
$$-\frac{(1- 9x)^3}{1+ (1- 9x)^2}$$
The chain rule is involved.

Take a simpler example, one whose integrand has an easy anti-derivative:

$\displaystyle \frac{d}{dx}\int_{1}^{1- 9x}\,t^2\,dt=\frac{d}{dx}\left(\frac{(1- 9x)^3}{3}-\frac{1^3}{3}\right) \\ \ \\ \ \\ \displaystyle \quad\quad\quad\quad\quad\quad\quad =3 \frac{(1- 9x)^2}{3} \frac{d}{dx} (1- 9x)-0 \\ \ \\ \ \\ \displaystyle \quad\quad\quad\quad\quad\quad\quad =-9(1- 9x)^2$