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zepp

derivative of an integral \[\frac{d}{dx}\int f(x^2)dx^2\]

  • one year ago
  • one year ago

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  1. zepp
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    i've never seen a \(\large dx^2\), anyone can explain to me what it is?

    • one year ago
  2. saifoo.khan
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    Zippy! Me meet again. ;D

    • one year ago
  3. zepp
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    ohai saif :D

    • one year ago
  4. saifoo.khan
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    not sure how to solve this. :/

    • one year ago
  5. AccessDenied
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    I recommend attempting the substitution: \(\sqrt{u} = x\). Just a hunch that it may turn out well (and not because I just explained the solution on irc chats). I think it may be helpful to our problem so that the f(x^2) dx^2 becomes f(u) du, etc...

    • one year ago
  6. AccessDenied
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    \[ \newcommand\dd[1]{\,\mathrm d#1} \newcommand\de[1]{\frac{\mathrm d}{\mathrm d#1}} \text{Let } \sqrt{u} = x \text{. Then, } \\ \; \frac{1}{2 \sqrt{u}} \dd{u} = \dd{x} \\ \begin{align*} \implies \de{x} \int f(x^2) \dd{x^2} &= \frac{\text{d}}{\left(\frac{1}{2\sqrt{u}}\right) \; \dd{u}} \int f(u) \; \dd{u} \\ &= 2\sqrt{u} \de{u} \int f(u) \; \dd{u} \\ &= 2x f(u) \\ &= 2x f(x^2) \\ \end{align*} \]

    • one year ago
  7. tkhunny
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    Another Way: \(\de{x}\int f(x^{2})\;dx^{2} = \de{x}\int f(x^{2})\cdot 2x\;dx = \de{x} \left(F(x^{2}) + C\right) = f(x^{2})\cdot 2x\) Same thing, really.

    • one year ago
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