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## Homework Statement:

- If $$u=\int_{x}^{(y-x)} \frac{sin t}{t} dt$$. Find ##\frac {\partial{u}}{\partial{x}}. \frac {\partial{u}}{\partial{y}}, \frac {\partial{y}}{\partial{x}}## at ##x=\frac{\pi}{2}; y=\pi##

## Relevant Equations:

- Leibniz Rule.

Because the limit of the integral is multi-variable, which is not explained at the ML Boas's example, I tried to start from the basic. First, I use:

$$\frac {dF}{dx}=f(x) \Rightarrow \int_a^b f(t) dt = F(b) - F(a)$$.

In my case now:

$$\int_{u(x)}^{v(x,y)} f(t) dt = F(v(x,y)) - F(u(x))$$

So,

$$\frac {d}{dx} \int_{u(x)}^{v(x,y)} f(t) dt = \frac {d}{dx} [F(v(x,y)) - F(u(x))]$$

$$\frac {d}{dx} \int_{u(x)}^{v(x,y)} f(t) dt = \frac {d}{dv} [F(v(x,y))]\frac {dv(x,y)}{dx} - \frac {d}{du} F(u(x)) \frac {du(x)}{dx}$$

$$\frac {d}{dx} \int_{u(x)}^{v(x,y)} f(t) dt = f(v) \frac {dv(x,y)}{dx} - f(u) \frac {du(x)}{dx}$$

where

$$dv(x,y)= \frac {\partial v}{\partial x} dx+\frac {\partial v}{\partial y} dy$$

So,

$$\frac {dv(x,y)}{dx}= \frac {\partial v}{\partial x} +\frac {\partial v}{\partial y} \frac {dy}{dx}$$

Now,

I put the detail on the question, where:

$$f(t) = \frac{\sin (t)}{t}$$

##v(x,y) = y - x/t## then ##dv = dy - dx##. So, ##\frac {\partial v}{\partial x}=-1## and ##\frac {\partial v}{\partial y} = 1##.

Now, with this, can I solve the problem? Thanks...

$$\frac {dF}{dx}=f(x) \Rightarrow \int_a^b f(t) dt = F(b) - F(a)$$.

In my case now:

$$\int_{u(x)}^{v(x,y)} f(t) dt = F(v(x,y)) - F(u(x))$$

So,

$$\frac {d}{dx} \int_{u(x)}^{v(x,y)} f(t) dt = \frac {d}{dx} [F(v(x,y)) - F(u(x))]$$

$$\frac {d}{dx} \int_{u(x)}^{v(x,y)} f(t) dt = \frac {d}{dv} [F(v(x,y))]\frac {dv(x,y)}{dx} - \frac {d}{du} F(u(x)) \frac {du(x)}{dx}$$

$$\frac {d}{dx} \int_{u(x)}^{v(x,y)} f(t) dt = f(v) \frac {dv(x,y)}{dx} - f(u) \frac {du(x)}{dx}$$

where

$$dv(x,y)= \frac {\partial v}{\partial x} dx+\frac {\partial v}{\partial y} dy$$

So,

$$\frac {dv(x,y)}{dx}= \frac {\partial v}{\partial x} +\frac {\partial v}{\partial y} \frac {dy}{dx}$$

Now,

I put the detail on the question, where:

$$f(t) = \frac{\sin (t)}{t}$$

##v(x,y) = y - x/t## then ##dv = dy - dx##. So, ##\frac {\partial v}{\partial x}=-1## and ##\frac {\partial v}{\partial y} = 1##.

Now, with this, can I solve the problem? Thanks...

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