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!
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?