# Thread: Find the derivative using Leibniz Rule

1. ## Find the derivative using Leibniz Rule

I found this question on Yahoo! Answers, and I am stumped (however, I would like to know the answer):

Let f(x,y) be a continuous function in R2. For t>0 let R_t = [0,t] x [0,2t] = { (x,y) | 0<=x<=t, 0<=y<=2t }
and define F(t) integral integral R_t of f(x,y)dA
Use Leibniz Rule to compute the derivative dF/dt.

How do I use Leibniz Rule? - Yahoo! Answers

Thanks, Charles

2. A lot, of course, depends on the function f(x,y).

Define $G(x, t)= \int_0^{2t} f(x,y)dy$.

Then $\frac{\partial G}{\partial x}= \int_0^{2t}\frac{\partial f}{\partial x} dy$.

Now, $F(t)= \int_0^t G(x,t)dx$ so that $\frac{dF}{dt}= G(t,t)+ \int_0^t\frac{\partial G}{\partial x}dx$.

$= \int_0^{2t}f(t, y)dy+ \int_0^t\int_0^{2t}\frac{\partial G}{\partial x}dydx$.

3. Wouldn't it be

$\displaystyle \int_0^{2t} f(t,y)dy + 2\int_0^t f(x,2t)dx$

4. Originally Posted by HallsofIvy
$= \int_0^{2t}f(t, y)dy+ \int_0^t\int_0^{2t}\frac{\partial G}{\partial x}dydx$.
Are you sure you didn't intend to write:

$= \int_0^{2t}f(t, y)dy+ \int_0^t\int_0^{2t}\frac{\partial f}{\partial x}dydx$

Anyway, your formula didn't work when I set f(x,y) = x sin y, although perhaps I made a mistake in my calculation. Are you sure your formula is correct?

Also, I don't understand why you have two terms in your result: G(t,t) and $\int_0^t\frac{\partial G}{\partial x}dx$

5. Originally Posted by Danny
Wouldn't it be

$\displaystyle \int_0^{2t} f(t,y)dy + 2\int_0^t f(x,2t)dx$
After carrying out the calculation, I concur with your result. Furthermore, f(x,y) need only be integrable, and continuity is more than sufficient. The Leibniz Rule was a red herring:

$\displaystyle [\int_0^{t+h} \int_0^{2t+2h} f(x,y) dydx - \int_0^t \int_0^{2t} f(x,y) dydx]/h$

$\displaystyle = [\int_0^t \int_0^{2t+2h} f(x,y) dydx + \int_t^{t+h} \int_0^{2t+2h} f(x,y) dydx - \int_0^t \int_0^{2t} f(x,y) dydx]/h$

$\displaystyle = [\int_0^t \int_0^{2t} f(x,y) dydx + \int_0^t \int_{2t}^{2t+2h} f(x,y) dydx$

$\displaystyle \hspace{10 mm} + \int_t^{t+h} \int_0^{2t} f(x,y) dydx + \int_t^{t+h} \int_{2t}^{2t+2h} f(x,y) dydx - \int_0^t \int_0^{2t} f(x,y) dydx]/h$

$\displaystyle = [\int_0^t \int_{2t}^{2t+2h} f(x,y) dydx + \int_t^{t+h} \int_0^{2t} f(x,y) dydx + \int_t^{t+h} \int_{2t}^{2t+2h} f(x,y) dydx]/h$

Approximating the integrals with (single term) Riemann sums, we get:

$\displaystyle \approx [{2h} \int_0^t f(x,2t) dx + h \int_0^{2t} f(t,y) dy + h \cdot 2h \cdot f(t,2t)]/h$

$\displaystyle \rightarrow 2 \int_0^t f(x,2t) dx + \int_0^{2t} f(t,y) dy, h \rightarrow 0$
$\displaystyle \therefore F'(t) = 2 \int_0^t f(x,2t) dx + \int_0^{2t} f(t,y) dy$