I'm working on a problem in calculus and am having difficulty with a specific function and its partial derivative. The function is defined as:

$F(u,v) = \int_{uv}^{u+v}e^{-(u-y)^2}dy$

I'm trying to calculate the partial derivative of this function with respect to $u$, evaluated at $u = 1$. Specifically, I'm looking to find $F'_u(1,v)$.

The correct answer is $(1-v) e^{-(1-v)^2}$

What I've Tried:

  • I understand that the process involves applying the Leibniz rule for differentiating under the integral sign.
  • I've attempted to differentiate inside the integral, but I'm unsure if I'm applying the rule correctly, especially with the changing limits of integration.

My Question:

Could someone help me understand the correct process for finding $F'_u(1,v)$? I'm particularly interested in:

  • The correct application of the Leibniz rule in this context.
  • Any special considerations that arise due to the limits of integration being functions of $u$ and $v$
  • Detailed steps or explanations would be greatly appreciated, as I'm looking to understand the process thoroughly.

Thank you in advance for your help!


1 Answer 1


Since you are only interested in the differential w.r.t. $u$, we may regard $v$ as constant. So we'll omit the $v$ completely. Temporarily, regard your $F$ as a function of three variables, and write $$H(a,b,u) = \int_{a}^b g(u,y) dy$$ where $g(u,y) = e^{-(u-y)^2}$. Then your function in question is $$F(u) = \int_{a(u)}^{b(u)} g(u,y)dy = H(a(u),b(u),u).$$ where $a(u) = uv$ and $b(u) = u+v.$ Note that \begin{align*} &H_a = -g(u,a)\\ &H_b = g(u,b)\\ &H_u = \int_a^b g_u(u,y) dy. \end{align*} By the chain rule, \begin{align*} F_u(u) = H_a(a(u),b(u),u) \cdot a_u(u) + H_b (a(u),b(u),u) \cdot b_u(u) + H_u(a(u),b(u),u). \end{align*} Can you take it from here?

  • $\begingroup$ I think I got you, but there must be a way to solve it without using a three variables function (because we haven't learned it, and it still was a question in the exam). $\endgroup$
    – liadperetz
    Dec 14, 2023 at 19:22
  • $\begingroup$ Another way would be to make the substitution $y = w+u$ to rewrite your integral as $F(u,v)= \int_{uv + u}^{2u+v} e^{-w^2} dw$. Perhaps this helps? $\endgroup$
    – Ssay
    Dec 14, 2023 at 19:29

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