1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivative of an integral question

  1. Jan 15, 2013 #1
    1. The problem statement, all variables and given/known data

    http://imgur.com/hCzGE

    2. Relevant equations



    3. 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: Jan 15, 2013
  2. jcsd
  3. Jan 15, 2013 #2
    What does the fundamental theorem state?
     
  4. Jan 15, 2013 #3
    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
     
  5. Jan 15, 2013 #4

    lurflurf

    User Avatar
    Homework Helper

    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]
     
  6. Jan 15, 2013 #5
    What are g and f in the problem?
     
  7. Jan 15, 2013 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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]
     
  8. Jan 15, 2013 #7

    lurflurf

    User Avatar
    Homework Helper

    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]
     
  9. Jan 15, 2013 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  10. Jan 15, 2013 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook