Given that: $T(x,y)=\ \int_{x-y}^{x+y} \frac{\sin(t)}{t}dt\ $, calculate: $\frac{\partial T}{\partial x}(\frac{\pi}{2}, - \frac{\pi}{2})$.

Question

Given that: $T(x,y)=\ \int_{x-y}^{x+y} \frac{\sin(t)}{t}dt\ $, How do I calculate: $\frac{\partial T}{\partial x}(\frac{\pi}{2}, - \frac{\pi}{2})$? I seriously have no direction for how to solve this question, I know that I need somehow to use the Chain Rule: $\frac{dF}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$ , but how do I start? any kind of directing/help would be appreciated.

Answer 1

We can think of: $$T(x,y) = \int_{a(x,y)}^{b(x,y)}f(t)\,{\rm d}t,$$ with $a(x,y) = x-y$, $b(x,y) = x+y$ and $f(t) = \sin(t)/t$. We have: $$T(x,y) = \int_0^{b(x,y)} f(t)\,{\rm d}t -\int_0^{a(x,y)}f(t)\,{\rm d}t, $$ so: $$\frac{\partial T}{\partial x}(x,y) = \frac{\partial b}{\partial x}(x,y) f(b(x,y)) - \frac{\partial a}{\partial x}(x,y) f(a(x,y)).$$

Now plug in the functions and the points and you are done.

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To understand this mess, think of the single variable case first: $$F(x) = \int_0^{g(x)}f(t)\,{\rm d}t \implies F'(x) = g'(x) f(g(x)).$$

Answer 2

$$\frac{\partial{T}}{\partial{x}}=\frac{\sin(x+y)}{x+y}-\frac{\sin(x-y)}{x-y}$$

Replacing $x=\frac{\pi}{2}$ and $y=-\frac{\pi}{2}$ we get

$$T_x(\frac{\pi}{2},-\frac{\pi}{2})=1$$

Using the fact that $\lim\limits_{x\to 0}\frac{\sin{x}}{x}=1$ and $\sin{\pi}=0$

I used the chain rule to get the initial derivative but it is trivial because $(x+y)_x=(x-y)_x=1$

Answer 3

here we go. what you have is $$T(x,y) = F(x+y) - F(x-y) \text{ where } F'(t) = \ \frac{\sin t}t.$$ now we can differentiate $T$ with respect to $x$ to get $$T_x = F'(x+y) \times 1 - F'(x-y) \times 1 =F'(\pi) - F'(0))= 1 -0= 1.$$