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Math Help - What is the derivative of an integral wrt the integrand?

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

    By the fundamental thm of calc, the derivative of a definite integral wrt to the upper limit is the integrand evaluated at the upper limit.

    \frac{d}{dy} \int^{y}_{0} f(x) dx = f(y)

    What if I integrate wrt to the integrand?

    \frac{d}{df(x)} \int^{y}_{0} f(x) dx = ?

    Thanks!
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by richardcherron View Post
    By the fundamental thm of calc, the derivative of a definite integral wrt to the upper limit is the integrand evaluated at the upper limit.

    \frac{d}{dy} \int^{y}_{0} f(x) dx = f(y)

    What if I integrate wrt to the integrand?

    \frac{d}{df(x)} \int^{y}_{0} f(x) dx = ?

    Thanks!
    \frac{d}{dy}\int_0^y{f}(x)dx

    =\frac{d}{dy}F(x)\Big|_0^y

    =\frac{d}{dy}[F(y)-F(0)]. But note that F(0) is constant which implies that

    \frac{d}{dy}[F(y)-F(0)]=\frac{d}{dy}[F(y)]+0=f(y)
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Oh, wait. I see what you're after.

    \frac{d}{df(x)}\int_0^y{f(x)dx}=\frac{d}{df(x)}[F(y)].

    So, if u=f(x), and v=F(y) we have

    \frac{dv}{du}
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    \frac{d}{dy}\int_0^y{f}(x)dx

    =\frac{d}{dy}F(x)\Big|_0^y

    =\frac{d}{dy}[F(y)-F(0)]. But note that F(0) is constant which implies that

    \frac{d}{dy}[F(y)-F(0)]=\frac{d}{dy}[F(y)]+0=f(y)
    I guess you misread, since he wants the derivative wrt f.


    Anyway, maybe you can write : \frac{d}{df({\color{red}y})}\int_0^y f(x) ~dx=\frac{dy}{df(y)} \cdot \frac{d}{dy} \int_0^y f(x) ~dx

    =\frac{1}{f'(y)} \cdot f(y)=\frac{f(y)}{f'(y)}
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