Derivative of an integral question

  • Thread starter hahaha158
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  • #1
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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
 
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  • #2
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What does the fundamental theorem state?
 
  • #3
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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
 
  • #4
lurflurf
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The question is find f'(x) for
[tex]\mathrm{f}(x)=\int^x_2 \! \left(\frac{1}{2}t^2-1 \right)^{15} \, \mathrm{dt}[/tex]

by the fundamental theorem of calculus

[tex]\dfrac{d}{dx}\int^x_a \! \mathrm{g}(t) \, \mathrm{dt}=g(x)[/tex]

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.

[tex]\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)[/tex]
 
  • #5
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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?
 
  • #6
HallsofIvy
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Homework Statement



http://imgur.com/hCzGE

Homework Equations





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
[tex]\int_{1- 9x}^1\frac{u^3}{1+u^2}du[/tex]

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

So that we can now apply the "fundamental theorem": replace each "u" in the integrand with 1- 9x. The derivative is
[tex]-\frac{(1- 9x)^3}{1+ (1- 9x)^2}[/tex]
 
  • #7
lurflurf
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Is this changing? First it was
[tex]\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}
[/tex]
 
  • #8
SammyS
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Is this changing? First it was
[tex]\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}
[/tex]
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.
 
  • #9
SammyS
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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
[tex]\int_{1- 9x}^1\frac{u^3}{1+u^2}du[/tex]

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

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

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

[itex]\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[/itex]
 

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