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Let $g \in C^1(\mathbb R)$ be a real valued function and $f$ defined by $$f(u,v,w) = \int_{u}^{v} g(w^2+\sqrt{s})\,\,ds$$ where $u,v,w \in \mathbb R$ and $u,v>0$. Find all partial derivatives.

I'm not sure how to attempt this problem. I assume if it were a function of two variables, say something like $$f(x,w) = \int_{0}^{x} g(w^2+\sqrt{s})\,\,ds$$ then for example the partial derivative with respect to $x$ would just be $g(w^2 + \sqrt{x})$ (is that true?).

Anyhow, some hint or strategy would be very welcomed.

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    $\begingroup$ I believe with respect to x it would be $g(w^2 +\sqrt{x})$. $\endgroup$ – Chinny84 Aug 5 '14 at 8:42
  • $\begingroup$ Yes, sorry, this is what I meant to write. I will edit it. $\endgroup$ – rehband Aug 5 '14 at 8:48
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    $\begingroup$ This function is not well defined. The domain is $\{ (u,v,w) \in \mathbb{R}^3 : u >0, v>0 \}$ because of that square root. $\endgroup$ – Crostul Aug 5 '14 at 8:54
  • $\begingroup$ Thanks, fantastic observation. I shall edit this as well. $\endgroup$ – rehband Aug 5 '14 at 8:56
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Call $h(s) = g(w^2-\sqrt{s})$. When you compute the partial derivatives with rispect to $u,v$ the variable $w$ is fixed, hence you can think $h$ as a function $\mathbb{R} \longrightarrow \mathbb{R}$.

So $$\frac{\partial}{\partial u} \int_u^v h(s) ds = - \frac{\partial}{\partial u} \int_v^u h(s) ds =- h(u)$$ since $v$ is considered a constant. In the same way you get $$ \frac{\partial}{\partial v} \int_u^v h(s) ds = h(v)$$ While for the third partial derivative you need to exchange the derivative with the integral sign, so you get $$\frac{\partial}{\partial w} \int_u^v g(w^2-\sqrt{s}) ds = \int_u^v \frac{\partial}{\partial w} g(w^2-\sqrt{s}) ds = \int_u^v g'(w^2-\sqrt{s}) 2w \ ds$$

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  • $\begingroup$ Excellent answer, thank you sir. $\endgroup$ – rehband Aug 5 '14 at 9:25
  • $\begingroup$ You are welcome. $\endgroup$ – Crostul Aug 5 '14 at 9:25

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