# How to Calculate the Partial Derivative F_u(1,v) of the Integral Function F(u,v)

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.

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?
• 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?