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Find the derivative of the integral from 0 to x^4 (sec(t)dt)

  1. Oct 11, 2006 #1
    find the derivative of the integral from 0 to x^4 (sec(t)dt)

    I get sec(x^4)(4x^3)

    I understand the procedure for solving this integral. However, I don't understand why exactly we need to use the chain rule in this situation.

    Why isn't the answer just sec(x^4)?

    Thank you!
  2. jcsd
  3. Oct 11, 2006 #2
    [tex] \frac{d}{dt} \int_{0}^{x^{4}} \sec t dt [/tex]. Basically you are getting a derivative. If its a derivative then you have to use the chain rule.
    Last edited: Oct 11, 2006
  4. Oct 11, 2006 #3
    Thanks for your reply. How about this problem

    d/dx integral 0 to ln(x) (sint + t^3/2)dt

    using the chain rule and ftc1 I get:

    u = ln(x) du/dx = 1/x

    d/du (sinu + u^3/2)du/dx

    sin(lnx)+(lnx)^3/2 (1/x)

    is this correct? Thanks again.
  5. Oct 11, 2006 #4
    [tex] \frac{d}{dx} \int_{0}^{\ln x} (\sin t + t^{\frac{3}{2}}) dt [/tex]. So the derivative so far, is [tex] \sin(\ln x) + (\ln x)^\frac{3}{2} dt [/tex]. So then it would be:

    [tex] \frac{\sin(\ln x)}{x} + \frac{(\ln x)^{\frac{3}{2}}}{x} [/tex].
    Last edited: Oct 11, 2006
  6. Oct 11, 2006 #5
    so my answer was off by not multiplying both parts by 1/x? Thanks.
    Last edited: Oct 11, 2006
  7. Oct 11, 2006 #6
    I took the derivative of u. i confirmed your answer.
  8. Oct 11, 2006 #7
    ok ok, thanks!!
  9. Oct 11, 2006 #8
    yeah, both parts need to be multiplied by [tex] \frac{1}{x} [/tex]. They both contain u and thus must contain du .
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