# partial derivative of an integral function

Given the function $f(x,y) = \int^x_y g(t) dt$, $g(t)$ continuous for all $t$.

How to evaluate the partial derivatives of $f$ ?

It is correct to do like this (by FTC) ?:

$f_x$ = $g(x)x'$.

$f_y = g(y)y'$.

Thanks!!!

• What do you understand by $x'$ and $y'$? – TZakrevskiy May 29 '17 at 12:27
• "Leibniz's formula" is a generalization of the "Fundamental Theorem of Calculus": – user247327 May 29 '17 at 12:29

Let $a$ be a real. then

$$f (x,y)=\int_a^x g (t)dt-\int_a^y g (t)dt$$

$$=G (x)-G (y)$$ with $$G(X)=\int_a^X g (t)dt$$ and $$G'(X)=g (X)$$ since $g$ is continuous at $\mathbb R$.

thus

$$f_x (x,y)=G'(x)=g (x)$$ $$f_y (x,y)=-G'(y)=-g (y)$$

"Leibniz's formula" is a generalization of the "Fundamental Theorem of Calculus": $$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x, t)dt= f(x, \beta(x))- f(x, \alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f(x, t)}{\partial x} dt$$

Here, $f(x,t)$ is a function of t only, the upper bound on the integral is just $x$ and the lower bound just $y$. So the derivative of the integral with respect to $x$ is $g(x)$ and the derivative with respect to y is $-g(y)$.