Find g'(x)

G(x)=9tcos(t^10)dt; intergral from [0 to sqrt(x)]

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asked by Andrew
  1. If the entire function g(x) is an integral with upper and lower limits, that makes it a definite integral. A definite integral always has a constant value.

    That would make g(x) a constant value. So what would its derivative be?

    (This might be incorrect if my interpretation of the question is inaccurate)

    posted by Arora
  2. using the 2nd FTC,

    g'(x) = 9√x cos(x^5) * 1/(2√x)

    If G(x) = ∫[a,u(x)] f(t) dt
    G'(x) = f(u(x)) * u'(x)

    This is just the chain rule, seen from the other side.

    posted by Steve
  3. Note that
    if F'(x) = f(x)
    and G(x) = ∫[u(x),v(x)] f(t) dt
    = F(v)-F(u)
    G'(x) = f(v)*v' - f(u)*u'

    If, as in your problem, the lower limit is a constant a, then F(a) is a constant, so F'(a) = 0 (that means that u is a constant, so u'=0 in the more general formula)

    posted by Steve

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