# Derivative of an integral question

• hahaha158
In summary, the fundamental theorem of calculus states that 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. In the given conversation, the question is about finding f'(x) using the fundamental theorem of calculus for the function f(x)= integral^x_2 of (1/2)t^2-1 raised to the 15th power. The differentiation cancels the integration and a generalization called Leibniz rule can be used. The chain rule is not needed in this problem.
hahaha158

## 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

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.

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

hahaha158 said:
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?

hahaha158 said:

## 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
$$-\frac{(1- 9x)^3}{1+ (1- 9x)^2}$$

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 said:
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.

HallsofIvy said:
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:

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

## 1. What is the derivative of an integral?

The derivative of an integral is a mathematical concept that represents the rate of change of the integral with respect to its independent variable. It is also known as the fundamental theorem of calculus and is represented by the notation "d/dx".

## 2. How is the derivative of an integral calculated?

The derivative of an integral is calculated by using the fundamental theorem of calculus, which states that the derivative of an integral is the function being integrated. This means that you can find the derivative of an integral by using the power rule, product rule, quotient rule, or chain rule depending on the function being integrated.

## 3. Why is the derivative of an integral important?

The derivative of an integral is important because it allows us to calculate the rate of change of a function, which is useful in many real-world applications. It also allows us to find the maximum and minimum values of a function, as well as the slope of a curve at a specific point.

## 4. Can the derivative of an integral be negative?

Yes, the derivative of an integral can be negative. This means that the function being integrated is decreasing at that particular point. The sign of the derivative depends on the direction of change of the function being integrated.

## 5. How is the derivative of an integral used in science?

The derivative of an integral is used in various fields of science, such as physics, chemistry, and engineering. It is used to calculate rates of change in physical systems, such as velocity and acceleration, as well as in the analysis of experimental data. It also plays a crucial role in the development of mathematical models and theories in various scientific disciplines.

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